Does that apply if the emitter is neither coherent nor opaque? If we were interested in gamma radiation, is the sum from each part of an infinite wall infinite (if we pretend the wall and the air are perfectly transparent to gamma rays)?

Note that in this case radiation is definitely incoherent. This rule applies to incoherent isotropic emission.

The coherent case is similar I believe, but because for an antenna (emitter) the field is usually tangent to the surface, you get some additional trigonometric factors (not purely solid angle). But by my calculations it's still 1/r^2.

Am I doing a calculation wrong then? For e.g. gravity towards an infinite plane, then the force is finite because components in opposite directions cancel out, but for radiation, then if you're 1m away from the wall, you could consider an infinite set of disjoint squares on the wall, each 2^n meters by 2^n meters and with the furthest point 2^(n+1) meters away from you. Since 2^2n / ((2^(n+1))^2) is 1/4 and an underestimate for the radiation you would receive from each section of wall, then the total radiation should be infinite.

Oh you might be right! I think what I wrote is incorrect in this case.

For gravity, it's clear the orientation of a small surface doesn't influence its gravitational force. In fact the source can be replaced by a point of equivalent mass.

For radiation, it seems the angle of a surface matters? For example, lambertian surfaces (everyday diffuse surfaces) do depend on their angle for total incoming radiation? Does only reflection exhibit lambertian behavior? There's something funny at work here.

In the lambertian case, there's a cosine term making your terms go to zero and sum converge.

When observing a small oblique (lambertian) surface, you get less radiation by a factor of cos(angle). I believe this is true for black body radiators as well (think a small glowing plate emitting red light). I'm not completely sure it would be true for x-ray and high energy radiation, and intuitively I suspect it indeed isn't. It seems like for high energy emissions, or generally isolated emissions (discounting opaqueness), only the total power is relevant like in gravity. I think self-absorption might be the culprit here.

If we say self-absorption (material absorption) in the high energy case is roughly negligible, then it becomes more strongly non-lambertian and the received power is simply a 1/r^2 integral (which would also explain why radiation gets stronger as it approaches the surface).

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Here's a simple proof of the solid angle formula I've found I believe is correct for lambertian surfaces: chop your object into infinitesimal pieces (limit goes to 0). For each piece, it tends to behave like a point, so we expect the radiation scaling with distance to be 1/r^2. But solid angle also scales as 1/r^2 for small enough pieces. (it's probably missing many technicalities to make the proof work, but I think it works). Note that for oblique surfaces their solid angle is a function of cos(angle), which follows Lambert's cosine law exactly as well.