I think this is an interesting and sort of deep question, so I'm going to answer it in part with the hope that my answer attracts even better answers.
I'll start with my first thought: surely there's no hope of formulating Green's theorem for an unbounded region, say the region $y > 0$. But then I thought about it for a moment, and observed that if you consider a smooth vector field $F(v)$ on the plane such that $F(v) \to 0$ rapidly as $v \to \infty$ then we can extend $F$ to the sphere by stereographic projection; this sends $y > 0$ to a hemisphere and the boundary curve $y = 0$ to the bounding great circle, and you can apply Stokes' theorem to this situation. Unwinding the calculations, this would give you a version of "Green's theorem" even for unbounded regions, albeit one that applies only to a certain class of vector field.
Then I thought about regions whose boundary is pathological, like the interior of the Koch snowflake. Here the boundary has infinite length, so surely there is no real hope of even defining the "boundary side" of Green's theorem. But then I noted that the Koch snowflake - like many pathological plane curves - has a very nice polygonal approximation, and it didn't sound insane that the boundary side could be defined as a limit of integrals over these approximations (again, maybe not for all vector fields). Sure enough, this has been worked outthis has been worked out, and there is indeed a version of Green's theorem for fractal boundaries.:
- Jenny Harrison and Alec Norton, The Gauss-Green theorem for fractal boundaries, Duke Math. J. 67 Number 3 (1992) pp. 575-588. doi:10.1215/S0012-7094-92-06724-X, author pdf.
There are other crazy things to try, like removing a non-measurable set from the plane or something. But Green's theorem (and its parent, the fundamental theorem of calculus) is based on a very resilient idea, something like "when you sum differences, things cancel". So in the spirit of the principle, "The fastest way to find something is to assert that it doesn't exist on the internet", I'll make a bold conjecture: Green's theorem can be generalized to any subset of the plane.