No, that's not it, because nothing holds |q| constant!

Instead, it's because of the electrical potential, which is the integral of the electric field from infinity. As you approach from infinity, if you approach a point, most of the charges on the object are "far away"; if you approach a big flat surface, though, a lot of them are "close". If the entire surface is going to have the same potential--if it's a conductor, any differences in potential on the surface will be evened out by charges moving until the outer surface is equipotential--that means that the pointy bits have to have a higher charge density locally to make up for the rest of the object being farther away. So instead of |q| = sigma*2*pi*r^2 for, say, a half-sphere, where sigma is the same for the object, sigma has to go up.

The overall effect often results in sigma looking roughly like 1/r, so that the electric field looks kind of like 1/r (not 1/r^2). But it depends on the configuration of the object; a tiny spike on a huge ball might have a tip with r a million times smaller than the ball, but have only a very slightly higher charge density because the electrical potential is dominated by the field emitted by the whole ball. (You'll still get breakdown at that spike, because breakdown will happen wherever the field exceeds the breakdown potential first, so the little boost from the spike will still mean that the spike wins--but if you have the spike on the side of a (U.S.-style) football, despite the spike being very sharp, you may get breakdown off the only-slightly-pointy ends of the ball before you get it off the thorn sticking out of the flat side.