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 flushed 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:
$OPQ$ 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." />