Let $K$ be the midpoint of $B M$, i.e., the centre of $\Gamma$. Notice that $A B \neq B C$ implies $K \neq O$. Clearly, the lines $O M$ and $O K$ are the perpendicular bisectors of $A C$ and $B M$, respectively. Therefore, $R$ is the intersection point of $P Q$ and $O K$.
Let $N$ be the second point of intersection of $\Gamma$ with the line $O M$. Since $B M$ is a diameter of $\Gamma$, the lines $B N$ and $A C$ are both perpendicular to $O M$. Hence $B N \parallel A C$, and it suffices to prove that $B N$ passes through $R$. Our plan for doing this is to interpret the lines $B N, O K$, and $P Q$ as the radical axes of three appropriate circles.
Let $\omega$ be the circle with diameter $B O$. Since $\angle B N O = \angle B K O = 90^\circ$, the points $N$ and $K$ lie on $\omega$.
Next we show that the points $O, K, P$, and $Q$ are concyclic. To this end, let $D$ and $E$ be the midpoints of $B C$ and $A B$, respectively. Clearly, $D$ and $E$ lie on the rays $O Q$ and $O P$, respectively. By our assumptions about the triangle $A B C$, the points $B, E, O, K$, and $D$ lie in this order on $\omega$. It follows that $\angle E O R = \angle E B K = \angle K B D = \angle K O D$, so the line $K O$ externally bisects the angle $P O Q$. Since the point $K$ is the centre of $\Gamma$, it also lies on the perpendicular bisector of $P Q$. So $K$ coincides with the midpoint of the $\operatorname{arc} P O Q$ of the circumcircle $\gamma$ of triangle $P O Q$.
Thus the lines $O K, B N$, and $P Q$ are pairwise radical axes of the circles $\omega, \gamma$, and $\Gamma$. Hence they are concurrent at $R$, as required.