Ok. Let's say $r \leq c$ and say $N = 2^r$.
Then the number of matrices with each column distinct is exactly $N(N-1)\cdots (N-c+1)$. This is an upper bound on the first question you asked.
Approximation time: suppose $r^2 \ll 2^c$ then almost all (asymptotically all) matrices will have distinct rows (birthday paradox) anyway. Since $r \geq c$, this condition will happen provided $r \gg 1$. Therefore, having distinct rows basically always happens, so the upper bound given above is essentially the truth [this can be made more rigorous and exact as desired].
If you want to look at the equivalence class thing, then just use the fact that most matrices will have only one automorphism anyway. So just divide the answer by $r! c!$.
Punchline: the thing is asymptotically easy, and the answer is more or less exactly what you'd think.