No.
For a simple counterexample, let's work in discrete time. Consider the following gambling strategy: start with \$0 and bet \$1 on a fair coin flip. If you win, you take your dollar and go home. If you lose, then bet \$100 on a second fair coin flip, and quit after that.
Think of $x = 50$. In order to have a chance to finish with more than \$50, you must lose a dollar on the first flip. So conditioned on finishing with more than \$50, the first round of the game is unfavorable - indeed a guaranteed loss.
Formally, we have $$\begin{align*} P(X_1 = X_2 = 1) &= 1/2 \\ P(X_1 = -1, X_2 = 99) &= 1/4 \\ P(X_1 = -1, X_2 = -101) &= 1/4 \end{align*}$$ but $Q(X_1 = -1, X_2 = 99) = 1$.
For a continuous time example, you could do something like the following: let $B_t$ be a standard Brownian motion, and $\tau_y$ the hitting time of level $y$. Take the stopping time $\tau$ defined by $$\tau = \begin{cases} \tau_1, & \text{if } \tau_1 \le 1 \\ 1, & \text{if } \tau_1 > 1 \text{ and } B_1 \ge -1 \\ \tau_{100}, & \text{otherwise}. \end{cases}$$ Then if $X_t = B_{t \wedge \tau}$ and $x=50$, we have $Q(X_1 < -1) = 1$.