Pick's Theorem on a triangular (or hex) grid

The short answer is that, no, there can be no formula for polygons with vertices in the hexagonal lattice in terms of just boundary and interior points. This is based on the fact that primitive triangles on this lattice--ones with no lattice points on their boundary (besides the vertices) or in the interior--can have different areas, whereas for the square lattice all primitive triangles have area $\frac{1}{2}$.

However, as Casebash has partly gotten at in his answer, you can approximate things well if you compute what, in the below paper, is called the "boundary characteristic" of the polygon, a number that is somewhat complicated to think to compute, but which gives a decent proxy for how many of each type of primitive triangle the polygon contains.

Kolodziejczyk has been the main one doing work on hexagonal lattice results of this type that I know; he's worth looking up for similar results. Ding Ren is another, and the older work of Grunbaum, etc., still bears on the problem.

"A Fast Pick-Type Approximation for the Area of H-Polygons," Ren, Kolodziejczyk, et al., American Mathematical Monthly, 1993.

Find the area of the polygon on the triangular grid in the usual way using Pick's formula, but multiply by $\sqrt{3}$ and divide by 2. This works because each square can be mapped to a parallelogram comprised of two equilateral triangles, and the area of the parallelogram is $\frac{\sqrt{3}}{2}$.

This is a very interesting question. I don't have a complete solution yet, but I did get some results. Consider an arbitrary distribution of points. Let $P(i,b)=i+\frac{b}{2}-1$ where $i$ is the number of internal points and $P$ is the number of boundary points. Let $P(A)=P(i_A,b_A)$, where $A$ is a simple polygon with all its vertices on the points. Wikipedia shows that $P(C)=P(A)+P(T)$ where $T$ is a triangle that shares a single edge with $A$ and $C$ a simple polygon formed by the union of $A$ and $T$. Since all simple polygons can be triangulated, $P(C)=sum P(t)$ for all $t$ in the triangulation of $C$.

The proof then has a second part that shows $P(t)$ equals the area for any triangle. So if we want to generalize it for other grids, we have to find a property equal to $P(T)$ for any triangle.