Assume that $\sigma_t$ is sufficiently larger than the "typical size" of the samples of $p_0$. Then we can make several heuristic observations:
$\mathbf{x} + \sigma_t \mathbf{z} \approx \sigma_t \mathbf{z}$ holds, since $\mathbf{x}$ is negligible compared to $\sigma_t \mathbf{z}$ with high probability.
On the other hand, the flow $f_t$ has the property that if $\mathbf{x}_t \sim p_t$, then $f_t(\mathbf{x}_t) \sim p_0$. Since $\mathbf{x} + \sigma_t \mathbf{z} \sim p_t$, we have $f_t(\mathbf{x} + \sigma_t \mathbf{z}) \sim p_0$.
From these two observations, we can expect that
$f_t(\mathbf{x} + \sigma_t \mathbf{z})$ is a sample of $p_0$ which is almost independent from $\mathbf{x}$ and is an almost deterministic function of $\mathbf{z}$. Consequently,
$$ p(f(\mathbf{x} + \sigma_t \mathbf{z}), \mathbf{z}) \approx p(f(\sigma_t \mathbf{z}), \mathbf{z}) $$
By the assumption, $f_t(\mathbf{x} + \sigma_t \mathbf{z})$ is also negligible compare to $\sigma_t \mathbf{z}$. Consequently, $\tilde{\mathbf{x}}_t \approx \sigma_t \mathbf{z}$.
As for the transportation cost, write $\mathbf{x}_t = \mathbf{x} + \sigma_t \mathbf{z}$ for simplicity and introduce the function
$$ I(t) = \mathbb{E}\bigl[ \| f_t(\mathbf{x}_t) - \mathbf{z}\|^2 \bigr], $$
which measures the transportation cost using the square distance cost function. Note that
$$ I(0) = \mathbb{E}\bigl[ \| \mathbf{x} - \mathbf{z}\|^2 \bigr], $$
hence OP's question can be rephrased to whether $ I(0) > I(t)$ or not. For this, the following observation comes handy:
Lemma. In addition to OP's setting, assume that $v_t(\mathbf{x}) = \mathbb{E}[\dot{\mathbf{x}}_t \mid \mathbf{x}_t = \mathbf{x}]$, which is the usual choice in the flow matching literature. Then $v_t$ generates the probability path $(p_t)$, and
\begin{align*}
v_t(\mathbf{x})
&= \dot{\sigma}_t \mathbb{E}[z \mid \mathbf{x}_t = \mathbf{x}]
= -\dot{\sigma}_t \sigma_t \nabla_\mathbf{x} \log p_t (\mathbf{x}).
\end{align*}
Using this, we can show that:
Claim. $I'(t) = -2\dot{\sigma}_t d + o(t)$ as $t \to 0^+$. In particular, if $\sigma_t$ is continuously differentiable with $\dot{\sigma}_0 > 0$, then $I(t)$ is decreasing for small $t$.
I am not sure if this trend will continue as $t$ grows, but my hunch is that the cost will indeed remain smaller. This is indeed true in the following special case:
Special Case. Assume $d = 1$ and $x \sim \mathcal{N}(0, 1)$. Then we can check that
$$ v_t(\mathbf{x}) = \frac{\dot{\sigma}_t \sigma_t}{1+\sigma_t^2} \mathbf{x}
\qquad\text{and}\qquad
f_t(\mathbf{x}) = \frac{1}{\sqrt{1+\sigma_t^2}} \mathbf{x}. $$
Using this, we can find an analytic formula for $I(t)$ :
$$ I(t) = 2 \left(1 - \frac{\sigma_t}{\sqrt{1+\smash[b]{\sigma_t^2}}} \right) $$
This is clearly decreasing in $t$. Also, it is easy to check that $I'(0) = -2\dot{\sigma}_0$, confirming the claim with $d = 1$.
Proof of Lemma. For any test function $\varphi \in C^{\infty}_c(\mathbb{R}^d)$, we evaluate the time-derivative of $\mathbb{E}[\varphi(\mathbf{x}_t)]$ in two ways. One one hand,
\begin{align*}
\frac{\partial}{\partial t} \mathbb{E}[\varphi(\mathbf{x}_t)]
&= \frac{\partial}{\partial t} \int_{\mathbb{R}^d} \varphi(\mathbf{x}) p_t(\mathbf{x}) \, \mathrm{d}\mathbf{x}
= \int_{\mathbb{R}^d} \varphi(\mathbf{x}) \frac{\partial}{\partial t} p_t(\mathbf{x}) \, \mathrm{d}\mathbf{x}.
\end{align*}
On the other hand, differentiating the random variable $\varphi(\mathbf{x}_t)$ directly and using the definition of $v_t$,
\begin{align*}
\frac{\partial}{\partial t} \mathbb{E}[\varphi(\mathbf{x}_t)]
&= \mathbb{E}[\nabla \varphi(\mathbf{x}_t) \cdot \dot{\mathbf{x}}_t] \\
&= \mathbb{E}[\nabla \varphi(\mathbf{x}_t) \cdot \mathbb{E}[\dot{\mathbf{x}}_t \mid \mathbf{x}_t]] \\
&= \mathbb{E}[\nabla \varphi(\mathbf{x}_t) \cdot v_t(\mathbf{x}_t)] \\
&= \int_{\mathbb{R}^d} \nabla \varphi(\mathbf{x}) \cdot v_t (\mathbf{x}) p_t(\mathbf{x}) \, \mathrm{d}\mathbf{x} \\
&= - \int_{\mathbb{R}^d} \varphi(\mathbf{x}) \nabla \cdot (v_t(\mathbf{x}) p_t(\mathbf{x})) \, \mathrm{d}\mathbf{x}
\end{align*}
where we used the law of iterated expectation in the second line and integration by parts in the last line. Now, since $\varphi$ is arbitrary, this shows that $v_t$ satisfies the continuity equation:
$$ \frac{\partial}{\partial t} p_t + \nabla \cdot (v_t p_t) = 0 $$
It is well-known that this implies that $v_t$ generates the probability path $p_t$. Next, invoking OP's settiing,
\begin{align*}
v_t(\mathbf{x})
&= \mathbb{E}[ \dot{\mathbf{x}}_t \mid \mathbf{x}_t = \mathbf{x}]
= \dot{\sigma}_t \mathbb{E}[\mathbf{z} \mid \mathbf{x}_t = \mathbf{x}] \\
&= \frac{\dot{\sigma}_t}{\sigma_t} \mathbb{E}[ \mathbf{x}_t - \mathbf{x}_0 \mid \mathbf{x}_t = \mathbf{x}]
= \frac{\dot{\sigma}_t}{\sigma_t} (\mathbf{x} - \mathbb{E}[ \mathbf{x}_0 \mid \mathbf{x}_t = \mathbf{x}] ).
\end{align*}
Since $\mathbf{x}_0$ is independent of the gaussian variable $\mathbf{z}$ and $\mathbf{x}_t = \mathbf{x}_0 + \sigma_t \mathbf{z}$, we can invoke Tweedie's formula to find:
$$ \mathbb{E}[ \mathbf{x}_0 \mid \mathbf{x}_t = \mathbf{x}] = \mathbf{x} + \sigma_t^2 \nabla \log p_t(\mathbf{x}). $$
Plugging this back, we obtain the desired identity.
Proof of Claim. Note that $f_t^{-1}(\mathbf{y})$ sends the initial point $\mathbf{y}$ at time $0$ along the vector field $v_t$ up to time $t$. Consequently, $f_t^{-1}$ is a flow with the corresponding vector field $v_t$, i.e.,
$$\frac{\partial}{\partial t} f_t^{-1}(\mathbf{y}) = v_t(f_t^{-1}(\mathbf{y})).$$
Using this observation, by differentiating both sides of the identity $\mathbf{x} = f_t(f_t^{-1}(\mathbf{x}))$ with respect to $t$, we find that the material derivative of $f_t$ is zero:
$$ \frac{\partial f_t}{\partial t} + \frac{\partial f_t}{\partial \mathbf{x}} v_t = 0 $$
Here, $\frac{\partial g}{\partial \mathbf{x}}$ stands for the Jacobian matrix of the multivariable function $g : \mathbb{R}^m \to \mathbb{R}^n$, and all the vectors are regarded as column vectors. Combining these altogether,
\begin{align*}
I'(t)
&= 2 \mathbb{E}\left[ \left\langle \frac{\partial}{\partial t}(f_t(\mathbf{x}_t)), f_t(\mathbf{x}_t) - \mathbf{z} \right\rangle \right] \\
&= 2 \mathbb{E}\left[ \left\langle \frac{\partial f_t}{\partial t}(\mathbf{x}_t) + \frac{\partial f_t(\mathbf{x}_t)}{\partial \mathbf{x}_t} \dot{\mathbf{x}}_t, f_t(\mathbf{x}_t) - z \right\rangle \right] \\
&= 2 \mathbb{E}\left[ \left\langle \frac{\partial f_t(\mathbf{x}_t)}{\partial \mathbf{x}_t} \left( \dot{\mathbf{x}}_t - v_t(\mathbf{x}_t) \right), f_t(\mathbf{x}_t) - z \right\rangle \right] \\
&= 2 \dot{\sigma}_t \mathbb{E}\left[ \left\langle \frac{\partial f_t(\mathbf{x}_t)}{\partial \mathbf{x}_t} \left( \mathbf{z} - \mathbb{E}[\mathbf{z} \mid \mathbf{x}_t] \right), f_t(\mathbf{x}_t) - z \right\rangle \right]
\end{align*}
Now let $\mathbf{w}_t = \mathbf{z} - \mathbb{E}[\mathbf{z} \mid \mathbf{x}_t] $. Then $\mathbb{E}[\mathbf{w}_t \mid \mathbf{x}_t] = 0$, hence $\mathbb{E}[ \langle \mathbf{w}_t, g(\mathbf{x}_t) \rangle ] = 0$ for essentially any function $g : \mathbb{R}^d \to \mathbb{R}^d$. Using this, we can further simplify the last line as:
\begin{align*}
I'(t)
&= - 2 \dot{\sigma}_t \mathbb{E}\left[ \left\langle \frac{\partial f_t(\mathbf{x}_t)}{\partial \mathbf{x}_t} \mathbf{w}_t, \mathbf{z} \right\rangle \right] \\
&= - 2 \dot{\sigma}_t \mathbb{E}\left[ \left\langle \frac{\partial f_t(\mathbf{x}_t)}{\partial \mathbf{x}_t} \mathbf{w}_t, \mathbf{w}_t \right\rangle \right]
\end{align*}
Now, when $t = 0$,
$f_0$ is the identity function, hence its Jacobian is the identity matrix: $\frac{\partial f_0}{\partial \mathbf{x}} = \mathbf{I}$.
$\mathbf{w}_0 = \mathbf{z} - \mathbb{E}[\mathbf{z} \mid x_0 = x] = \mathbf{z}$, where the last equality follows from the independence between $\mathbf{z}$ and $\mathbf{x}$.
Consequently,
$$ I'(0) = -2\dot{\sigma}_0 \mathbb{E}[\|\mathbf{z}\|^2] = -2\dot{\sigma}_0 d, $$
and the conclusion follows from the continuity of $I'(t)$.
