Write the governing SDEs as:
$$
\begin{cases}
d v_t = \alpha v_t dB_t^1 \\
d X_t = \left( 1 - \frac{\beta}{(1-\beta) (1-\rho^2)} \right) dt + 2 \sqrt{X_t} \left(\rho dB_t^1 + \sqrt{1-\rho^2} d B_t^2 \right)
\end{cases}
$$
where $B_t^1$ and $B_t^2$ are iid standard Brownian motions. As the OP suggests, one can solve for $v_t$ independently of $X_t$. In particular,
$$
v_t = v_0 \exp\left( -\frac{\alpha^2}{2} t + \alpha B_t^1 \right)
$$
It immediately follows from this expression that $v_t \to 0$ almost surely as $t \to \infty$, and hence, that the quadratic variation of $v_t$ almost surely asymptotes to a finite value. Thus, the proposed time change:
$$
\tau_t = (1-\rho^2) \int_0^t v_s^2 ds
$$
is **not** invertible. This lack of invertibility has practical consequences, among which there seem to be no explicit formulas for the expected values requested by the OP, and so one probably has to resort to Monte-Carlo.

To be specific, consider
$$
\begin{aligned}
\mathbb{E} X_{\tau_t} &= X_0 +
\left( 1 - \frac{\beta}{(1-\beta) (1-\rho^2)} \right) \mathbb{E} \tau_t \\
&+
2 \rho \mathbb{E} \int_0^{\tau_t} \sqrt{X_s} d B_s^1
+
2 \sqrt{1-\rho^2} \mathbb{E} \int_0^{\tau_t} \sqrt{X_s} d B_s^2
\end{aligned}
$$
The term involving $\mathbb{E} \tau_t$ is trivial to evaluate analytically. Unfortunately, the stochastic integrals appearing in this expression are not martingales because of the lack of invertibility in the proposed time change. Thus, as I said, one probably has to resort to Monte-Carlo to compute this quantity, and ditto for the variance of $X_{\tau_t}$. (The problem is naturally a bit more involved for the variance.)

Of course, to make the time change invertible one can, e.g., add a constant drift term $\mu dt$ to the SDE for $v_t$ with $\mu>-\alpha^2 / 2$.