This is a question from classical probability theory. I suppose $a_i$ are real (and $\alpha_i \ge 0$), though it's also possible to consider the complex case. The probability of obtaining $|0\rangle$ from the measurement of $\alpha_{0} |0\rangle + \alpha_{1}|1\rangle$ is $\alpha_0^2$. This probability is in fact a conditional probability $P(|0\rangle ~~\vert~~|\phi\rangle = \alpha_{0} |0\rangle + \alpha_{1}|1\rangle )$ of obtaining $|0\rangle$ from $|\phi\rangle$ where the condition is $|\phi\rangle = \alpha_{0} |0\rangle + \alpha_{1}|1\rangle$. In a discrete case (if $\alpha_0$ has only finite number of possible values) to calculate overall probability of obtaining $|0\rangle$ from $|\phi\rangle$ we could use the law of total probability
$$
P(|0\rangle) = \sum_{a_0} P(|0\rangle ~~\vert~~|\phi\rangle = \alpha_{0} |0\rangle + \alpha_{1}|1\rangle )P(|\phi\rangle = \alpha_{0} |0\rangle + \alpha_{1}|1\rangle)
$$
But we have a continuous case here with $P(|\phi\rangle = \alpha_{0} |0\rangle + \alpha_{1}|1\rangle) = 0$.
So we need to take the integral. There is a theory of how to calculate such probabilities with integrals, but I think in our case it is the correct way just to assume that $\alpha_0^2$ is uniformly distributed on $\{\frac{k}{n}, k=1..n\}$, calculate probability $P_n(|0\rangle)$ and then take the limit for $n \rightarrow \infty$.
$$
P_n(|0\rangle) = \sum_{k=1}^n P(|0\rangle ~~\vert~~|\phi\rangle = \sqrt{\frac{k}{n}} |0\rangle + \sqrt{\frac{n-k}{n}}|1\rangle )P(|\phi\rangle = \sqrt{\frac{k}{n}} |0\rangle + \sqrt{\frac{n-k}{n}}|1\rangle) =
$$
$$
= \sum_{k=1}^n \frac{k}{n} \cdot \frac{1}{n} = \sum_{k=1}^n \frac{k}{n^2} = \frac{n(n+1)}{2}\cdot\frac{1}{n^2} = \frac{n+1}{2n}
$$
The limit is $\frac{1}{2}$, which coincides with the probability of obtaining $|0\rangle$ from measuring $|\psi\rangle = |+\rangle$
Remark
Note that I assumed that you are given a lot of copies of $|\psi\rangle$ or a lot of samples of $|\phi\rangle$ (with varying $\alpha_0$) and you are asked to distinguish the case. If you are given only one copy of $|\psi\rangle$ or $|\phi\rangle$, then you can't distinguish it (with certainty) simply because the measurement of $|+\rangle$ will give random result.
Update
Another approach is to calculate the expected density matrix of $|\phi\rangle$:
$$
E(|\phi\rangle\langle\phi|) = \int_0^1(\sqrt{x_0}|0\rangle+\sqrt{1-x_0}|1\rangle)(\sqrt{x_0}\langle 0|+\sqrt{1-x_0}\langle 1|)dx_0 =
$$
$$
= \int_0^1x_0dx_0 \cdot |0\rangle\langle 0| + \int_0^1(1-x_0)dx_0 \cdot |1\rangle\langle 1| + $$
$$ + \int_0^1\sqrt{x_0(1-x_0)}dx_0 \cdot |0\rangle\langle 1| + \int_0^1\sqrt{(1-x_0)x_0}dx_0 \cdot |1\rangle\langle 0| =
$$
$$
= \frac{1}{2}(|0\rangle\langle 0| + |1\rangle\langle 1|) + \frac{\pi}{8}(|0\rangle\langle 1| + |1\rangle\langle 0|)
$$
Hence the overall probability of obtaining $|0\rangle$ from measuring $|\phi\rangle$ in computational basis is $\text{Tr}(E(|\phi\rangle\langle\phi|) \cdot |0\rangle\langle 0|) = \frac{1}{2}$.
Here we can see the difference with the complex case, since the expected density matrix will be $\frac{1}{2}I$.
