Consider a function $F$ on the half space $\{(x,y,z)|z>0\}$. If $F$ is analytic, it is straightforward to show that
A) The integral of $F$ over the hemisphere $(x-x_0)^2 + (y-y_0)^2 + z^2 = R^2$ vanishes for all $x_0$, $y_0$, and $R$.
implies
B) $F = 0$ everywhere
My question is whether this is known to be true for any continuous function (or under some less restrictive conditions).
As @alvarezpaiva notes, this is a question about hyperbolic hyperplane (Radon) transform. This has been studied. See, e.g., Kurusa, the Radon transform in hyperbolic space [Geometriae Dedicata, 1991] (available for free, thanks to Springer). and references therein (Kurusa inverts it on fairly natural subspaces of $L^2,$ I am not sure if his results are optimal.
I think that assuming that $F$ is continuous and decaying sufficiently fast at infinity is enough (although maybe the theorem is true in greater generality).
Here is a (sketchy) proof: The half-spheres over which you integrate are precisely the hyperplanes in hyperbolic space (in the half-space model). You're missing a few: the vertical hyperplanes, but those are really very few by comparison. Your problem is then to show that the (hyperplane) Radon transform on hyperbolic space is injective. I'm sure you'll find this in Helgason book(s).
Remark. I thought the following idea would reduce everything to the usual Radon transform, however there is a gap.
Among the models of hyperbolic space you have one in which the space is the interior of a ball in $\mathbb{R^3}$ and the hyperplanes are the intersections of this ball with hyperplanes in $\mathbb{R^3}$. If you switch to this model, then the problem becomes one for the standard Radon transform whose injectivity is quite classical (just use the relation between Radon transform and Fourier transform).
The gap: the Cayley-Klein model is not conformal to the half-space model. The resulting transform is not really the Radon transform on Euclidean three space (modulo multiplication of the function by the conformal factor), but a sort of weighted Radon transform. Is there a way to work around this? *What are the results for this sort of weighted Radon transforms?*
