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 $\mathrm{GL}(2,q)$ is needed. This takes a fairly different approach from Dickson which is based on the geometric actions of $\mathrm{PSL}(2,\mathbb{C})$, and instead uses a more module-theoretic approach, some of which goes back Suprunenko especially as carried on by Short. The classes of $\mathrm{PGL}(2,q)$ split in somewhat unusual and hard to control ways (I found the dihedrals to be a nightmare), but subgroups of $\mathrm{GL}(2,q)$, like subgroups of $\mathrm{Sym}(n)$, 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
- $\tau(q^2−1) − \tau(q−1)$ classes of irreducible, but not absolutely irreducible abelian subgroups (Singer)
- $\tau(q−1) \log_p q$ classes of indecomposable, but reducible abelian groups (central*unipotent)
Here $a$,$b$ are (weakly) multiplicative functions with values on prime powers:
- $a(p^e) = (p^{e+2} + p^{e+1} + 1 + 2e − 3p − 2ep) / (p−1)^2$
- $b(2^e) = 2e^2−2e+3$
- $b(p^e) = (e+1)^2$, $p$ odd
These functions are fairly natural: $a(n)$ counts the number of subgroups of $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$, 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 $\mathrm{GL}(2,q)$ for $q=p^e$, $p \geq 5$, 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 $\mathrm{GL}(2,q)$ 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 "$\mathrm{PGL}$" part, Dickson's method appears to just work.
