Note that there is a dilation centered at $A$ with ratio $2$ sending triangle $AMN$ to $ABC$. Hence the circumcircles of triangles $ABC$ and $AMN$ are tangent at $A$. Denote their common tangent at $A$ by $\ell$; we note that $\ell$ is the radical axis of these two circles. We now have a key lemma.

Lemma 1. Points $M$, $N$, $P$, and $Q$ lie on a circle.

Proof. Let $B_1$ be the point diametrically opposite $B$ on $\omega$ so that $BB_1$ is a diameter of $\omega$. Hence $B_1C \perp BC$ and $B_1A \perp BA$, meaning that $B_1C \parallel AH$ and $B_1A \parallel CH$. Thus $AB_1CH$ is a parallelogram. In parallelogram $AB_1CH$, diagonal $AC$ and $B_1H$ bisect each other, hence $N$ is the midpoint of $B_1H$. Note that $B_1H \cdot HQ$ is the power of $H$ with respect to $\omega$, so
$$NH \cdot HQ = \frac{1}{2} B_1H \cdot HQ$$
is equal to half of the power of $H$ with respect to $\omega$. In an analogous manner, we can show that $MH \cdot HP$ is also equal to half of the power of $H$ with respect to $\omega$. We conclude that $MH \cdot HP = NH \cdot HQ$, so $MNPQ$ is cyclic by power of a point. $\square$

Let us now finish the proof. By Lemma 1, we see that $MN$ is the radical axis of the circumcircles of $AMN$ and $MNPQ$, and that $PQ$ is the radical axis of the circumcircles of $ABC$ and $MNPQ$. Hence $R$ is the radical center of the three circumcircles. Consequently, we conclude that $R$ lies on $\ell$ and $AR \perp OA$, as desired.