Black Body Golf Balls
Consider a smooth spherical shell. The outside of the shell admits no energy transfer via radiation or any other means, so that the shell and its interior form an isolated system. The shell is at temperature $T$. Its inside surface is a perfect black body.
Next put a smooth solid sphere, also a perfect black body, in the center of the shell, also at temperature $T$. It will absorb radiation from the inside surface of the shell and will also emit radiation, all of which is absorbed by the shell. Because the shell and ball are at the same temperature, they cannot exchange net heat, so the sphere absorbs and emits radiation at the same rate.
Next replace the sphere with a perfect black body golf ball of equal cross section to the sphere. This ball absorbs the same amount of radiation from the shell as the sphere did simply because it absorbs all the radiation that hits it. The golf ball, like the solid sphere, must emit and absorb the same amount of radiation because otherwise heat would flow between bodies of equal temperature. Therefore, the golf ball emits the same amount of radiation as the sphere.
If you start with a sphere and put dimples in it, you've actually reduced the cross-section a bit because the caps of the dimples are gone, so such a dimpled sphere would radiate slightly less than before.
No, the golf ball would not radiate more. At http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law, after deriving the Stefan-Boltzmann law, it says:
Finally, this proof started out only considering a small flat surface. However, any differentiable surface can be approximated by a bunch of small flat surfaces. So long as the geometry of the surface does not cause the blackbody to reabsorb its own radiation, the total energy radiated is just the sum of the energies radiated by each surface; and the total surface area is just the sum of the areas of each surface—so this law holds for all convex blackbodies, too, so long as the surface has the same temperature throughout. The law extends to radiation from non-convex bodies by using the fact that the convex hull of a black body radiates as though it were itself a black body.
Thus, the convex hull of the golf ball is the surface that is relevant for calculating the total radiated energy. In other words, a sphere (EDIT: actually, the convex hull of the golf ball is the golf ball with the dimples replaced by discs, which has a surface area that is less than a sphere of the same radius)
Great question - harkens back to the days when we discussed why a tiny hole in a hot cavity is the perfect radiator and such things. Makes you think, which is what's great about it!
Would this increased surface area allow a black body golf ball to radiate more than its smooth counterpart?
The correct answer is no, it would radiate a tiny bit less than for a perfectly spherical golf ball, depending on how far away you are. If Bubba Watson is about to strike a ball on the tee directly your way, you want to be at a safe distance. How far? Let's say at least half a kilometer, i.e., 500m.
A good reference for understanding blackbody radiation and associated issues (like the "principle of detailed balance" and "Lambert's Law", which is essentially $\cos(\theta)$) is your trusty copy of Reif's "Fundamentals of Statistical and Thermal Physics".
Imagine something that's not quite a sphere - let's say two spheres connected by a long ultra-thin (massless) wire, just enough to call the arrangement a single body. Will this combination radiate more than a single sphere? Of course! But not along the line joining the two spheres, since one will be occluded by the other.
So the shape of the radiator does matter. Now let's consider the golf ball. The dimples are concave, so let's start with a single small dimple at the north pole of an otherwise smoothly spherical ball. If you are on the $z$-axis, directly above the north pole, you see the same cross-section ($\pi R^2$, where $R$ is the radius of the ball) radiating towards you. To get the exact amount of radiation you have to integrate over the surface facing you, using "Lambert's Law" alluded to above, and get the amount of radiation. The only difference from the unblemished smooth spherical golf ball case is that the dented part of the surface is slightly farther away from the observer.
To do the integral over the surface for the amount of radiation headed in a particular direction (towards the observer) you have to multiply the surface area element by $\cos(\theta)$, which is the dot product of the unit direction vector $\hat n$ and the area element $d\vec A$. This is what turns the spherical area of $2\pi R^2$ into a projected area of $\pi R^2$.
The only difference for the case of a real golf ball, and an observer viewing it from a random direction, is that the radiating surfaces are slightly farther away due to the dimples. But this is a tiny effect. If the dimples have an effective depth $d$, then for an observer at a distance $L$ large compared to the size of the ball the reduction in radiation will be the 35% dimple area estimate times $(1-d^2/L^2)$. If we estimate the effective depth to be half a millimeter, then at 500m we get roughly 1/3 of $10^{-12}$ less radiation compared to the perfectly spherical ball - a truly tiny effect.