It is a very good question with a subtle answer! Let me give you a short answer and then recommend a good paper that you can take a look.
The time evolution of a squeezed states defined as you mentioned is related to the way that the squeezing is produced. The squeezed states are produced in nonlinear processes in which a "classical" electromagnetic field drives a nonlinear medium. In the interaction picture the corresponding Hamiltonian for one such nonlinear optical process, that of parametric down-conversion can be written as
\begin{equation}
\hat{H}_I = [\epsilon (a^{\dagger})^2+\epsilon^* (a)^2]
\end{equation}
The quantity $\epsilon$ includes the amplitude of the driving field as well as the second-order susceptibility for the down-conversion. The operator $\hat{U}(t) = \exp(-i\hat{H}t/\hbar$) describing the time evolution of the single-mode under consideration can be written as
\begin{equation}
\hat{S}(\xi) = \exp\left[\xi \frac{(\hat{a}^{\dagger 2})}{2}-\xi^* \frac{(\hat{a}^{2})}{2} \right]
\end{equation} where $\xi = -i\epsilon t$. It is this quantaty $\xi$ (also given by $\xi = r \exp(i\theta)$) which will determine the size of the squeezing and will depend on the interaction time (essentially the lenght of the nonlinear medium). This idea can also be generalized to the case of a squeezed coherent states. Finally, the time dependence of the canonical operators $\hat{x}$ and $\hat{p}$ will not be explicity as in the past case. If you take a look on the quadrature operators you will came to the conclusion that if the evolution continues for time t, we will have
\begin{align}
\hat{X}_1(t) &= \hat{S}^\dagger(\xi) \, \hat{X}_1(0) \, \hat{S}(\xi) = \hat{X}_1(0) e^{-r} \\ \hat{X}_2(t) &= \hat{S}^\dagger(\xi) \, \hat{X}_2(0) \, \hat{S}(\xi) = \hat{X}_2(0) e^{r}
\end{align}
For a good discussion and detailed discussion about these issues, please check the following paper: https://arxiv.org/pdf/1401.4118.pdf;
