Thanks very much to domotorp for resolving this with an elegant symmetry based argument.  Since I could not immediately see how the crucial step - the emptyness of ORP - follows from symmetry, I've fleshed that portion out here:

We need the following result, which is simple enough that I do not include a proof:

Let $A \subseteq \mathbb{R}^2$ and suppose $T : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is a 1-1 and onto transformation that preserves $\mathbb{Z}^2$, that is $T(\mathbb{Z}^2)=\mathbb{Z}^2$.  Then $A$ contains a ``lattice point'' (by which we mean an element of $\mathbb{Z}^2$) if and only if $T(A)$ does.

Let us apply the above with $A=OPQ$ and $T$ denoting a reflection about the horizontal line whose $y$-coordinate is the average of the $y$-coordinates of $O$ and $P$ followed by a reflection about the vertical line whose $x$-coordinate is the average of the $x$-coordinates of $O$ and $P$.  Since both reflections are lattice preserving operations and since $T(OPQ) = ORP$, it follows that $OPR$ is empty if $OPQ$ is, as claimed.

And here's a Figure:

![Triangle $OPQ$ contains no interior lattice points when $P$ is the lattice point making the shallowest angle with $\ell$ from below.  By symmetry, in this case triangle $ORP$ also contains no interior lattice points.][1]


  [1]: https://i.sstatic.net/vrePS.png