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This seems to be true for the following reason: Suppose that $Y_0$ is close to zero, in which case $Y_t$ is well approximated in a strong sense by the process $\tilde Y_t$ which satisfies the linear SDE: $$ d \tilde Y_t = X_t \tilde Y_t dt + X_t \tilde Y_t d V_t \;. $$ This linear SDE has the pathwise solution: $$ \tilde Y_t = \tilde Y_0 e^{ \int_0^t X_s dV_s + \int_0^t ( X_s - \frac{1}{2} X_s^2) ds } \;. $$ Since $$ \frac{1}{t} \left( \int_0^t X_s dV_s + \int_0^t ( X_s - \frac{1}{2} X_s^2) ds\right) \to - \frac{1}{4} \quad \text{as $t \to \infty$ } $$ it follows that $\tilde Y_t \to 0$ as $t \to \infty$, and by approximation, that $Y_t \to 0$ too.