This question is reasonably hard, but important. A very clear and explicit answer is given in:
Flannery, D. L.; O'Brien, E. A. "Linear groups of small degree over finite fields." Internat. J. Algebra Comput. 15 (2005), no. 3, MR2151423 doi:10.1142/S0218196705002426
This has applications to primitive, solvable, linear groups of prime-squared degree and many other problems where an explicit knowledge of the subgroups of GL(2,q) is needed. This takes a fairly different approach from Dickson which is based on the geometric actions of PSL(2,C), and instead uses a more module-theoretic approach, some of which goes back Suprunenko especially as carried on by Short. The classes of PGL(2,q) split in somewhat unusual and hard to control ways (I found the dihedrals to be a nightmare), but subgroups of GL(2,q), like subgroups of Sym(n), can be classified by their action on the natural space.
This gives a simple formula for the number of conjugacy classes of abelian groups:
- (a(q−1)−b(q−1))/2 + b(q−1) classes of diagonal subgroups, a,b defined below
- τ(q2−1) − τ(q−1) classes of irreducible, but not absolutely irreducible abelian subgroups (Singer)
- τ(q−1)log(q,p) classes of indecomposable, but reducible abelian groups (centralunipotent)
Here a,b are (weakly) multiplicative functions with values on prime powers:
- a(pe) = (p(e+2) + p(e+1) + 1 + 2e − 3p − 2ep) / (p−1)2
- b(2e) = 2e2−2e+3
- b(pe) = (e+1)2, p odd
These functions are fairly natural: a(n) counts the number of subgroups of Z/nZ×Z/nZ, and b(n) counts the number of those subgroups left invariant by a coordinate swap.
I am still working through the details of the non-abelian groups, but do not foresee any problems. The paper handles GL(2,q) for q=pe, p ≥ 5, but for the most part I only need e=1, and the omissions in the paper are not too serious even for p=2,3.
A reducible subgroup of GL(2,q) must be abelian, and so the next case are the non-abelian imprimitive groups, all of which must be monomial and so have a clear list of representatives. The primitive linear groups seem to be messier in the details, but as one can more clearly distinguish the "Z" from the "PGL" part, Dickson's method appears to just work.