> Is there a chance that the r in the equation in this case would actually represent the distance between the surface of the sphere and the point, rather than the center?

The charges for a conducting sphere distribute themselves uniformly on the surface of the sphere. Each little element of charge on the surface contributes an infinitesimal amount of the final electric field. To calculate the final field, you need to (vector) integrate the contributions from all these elements. So you can't just use kq/r^2 but with r the "altitude" of the test charge.

If you do this calculation, you'll find that it actually can apply E = kq/r^(2) with r the distance to the centre of the sphere -- the uneven contributions cancel out. From the outside, a spherical shell of uniform charge looks exactly like a point charge at its centre.

This is actually true for any company spherical charge distribution, and you can prove it very elegantly using Gauss's law.