Let us assume that $f: \mathbb{R}^3 \to \mathbb{R}$ is sufficiently nice and that the following integral is convergent.
$$ \int \limits_{V \subseteq \mathbb{R}^3} f(x) dx $$
Can this integral be always computed as a surface integral over $\partial V$? My reasoning is that following the answer to this question we have that if $f$ is sufficiently nice then there is a solution to a Poisson equation:
$$ \nabla^2 \varphi = f $$
In that case, if we define $g = \nabla \varphi$ then we have that by Stokes theorem:
$$ \int \limits_{V \subseteq \mathbb{R}^3} f(x) dx = \int \limits_{V \subseteq \mathbb{R}^3} \nabla \cdot g(x)dx = \int \limits_{\partial V \subseteq \mathbb{R}^3} g(x) dx .$$
Sure. I mean, you could even just trivially define $$g = \frac{\int_{V} f(x)\,dV}{\int_{\partial V} 1\,dA}$$ and $f$ doesn't even need to be "nice," beyond integrable.
Your construction works as well, though you need to be careful with boundary conditions (in particular Dirichlet boundary conditions will work, but Neumann will not!)
Of course, usually the problem isn't existence of $g$; it's existence of a computationally convenient $g$ so that the surface integral is simpler to calculate than the original volume integral. Solving the Poisson equation is often no easier than just integrating the original function.