No. Consider any state with a momentum wavefunction symmetric about zero. It's position-space and momentum-space norm-squared probability distributions are not changed by time-reversal, even though the wavefunction clearly is.
Here is an explicit example. Take the four Gaussian wavepacket of mean positions $x_0$ or $-x_0$, mean momenta $p_0$ or $-p_0$, and spatial spread $\sigma$:
$\psi_{(\pm_x , \pm_p)} (x) \propto e^{-(x \mp_x x_0))^2/4\sigma^2 - i x (\pm_p p_0) }$.
(Here, $\pm_x$ and $\pm_p$ are two binary variables taking the values of plus or minus. $\mp_x$ and $\mp_p$ are their opposites.) Now consider the two superpositions
$\phi_{\mathrm{away}} = \psi_{(+,+)} + \psi_{(-,-)} $,
$\phi_{\mathrm{toward}} = \psi_{(+,-)} + \psi_{(-,+)} $.
$\phi_{\mathrm{away}}$ is the superposition of two wavepackets separated by $2 x_0$ traveling away from each other with relative speed $2 p_0/m$. $\phi_{\mathrm{toward}}$ is the same with the packets traveling toward each other. One can check that $\vert \phi_{\mathrm{toward}} (x)\vert^2 = \vert \phi_{\mathrm{away}} (x) \vert^2$ and $\vert \tilde{\phi}_{\mathrm{toward}} (p) \vert^2 = \vert\tilde{\phi}_{\mathrm{away}} (p) \vert^2$.
I do not know if there are examples other than with time-reversal.