With the current hard and software the set cover problem with 138,639,780 sets is too large to be handled.
The idea is to group Hcosets together into bigger sets, so that the set cover problem has a much smaller size. This can be done in a in a natural way by introducing two new subgroups K and T, closely related to the subgroups H and Q.
Subgroup 
Color Pattern 
Number of Elements 
Symmetry Type
Symmetries 
H 

H = 8! * 8! * 4! / 2
= 19,508,428,800 

Q 

Q = 4! * 4! * 4! * 4! * 4! / 2
= 3,981,312
= H / 4900 

K 

K = 8! * 8! * 4! * 3^{7} / 2
= 42,664,933,785,600
= H * 2187 
D4h
16 
T 

T = 4! * 4! * 4! * 8! * 3^{7} / 2
= 609,499,054,080
= K / 70

Oh
48 
The subgroup K can be described as the union of 2187 cosets of H having all possible 2187 corner orientations while the edges have the same properties as in the subgroup H.
The subgroup T derives likewise from the subgroup Q by dropping the restrictions on the corners. It is the union of C(8,4) * 2187 cosets of Q.

Now we list the properties of the corresponding cosets
Cosets of 
Coset Example 
Number of Cosets 
Symmetryreduced number 
H 

[G:H] =
C(12,4) * 3^{7 }* 2^{11}
= 2,217,093,120 
138,639,780 
Q 

[G:Q] =
C(12,4) *
C(8,4)^{2} * 3^{7} * 2^{11}
= 10,863,756,288,000 =
[G:H] * 4900 
226,331,259,372 
K 

= 1,013,760
=[G:H] / 2187 
64,430 
T 

[G:T] =
C(12,4) * C(8,4) *
2^{11}
= 70,963,200
= [G:K] * 70

1,482,170 
There are C(12,4) ways to position the redgreen edges and 2^11 ways to flip the 12 edges which gives the number 1,013,760 of different Kcosets. The number of essentially different cosets reduced by the 16 D4h symmetries can be shown to be 64,430.
There are C(12,4) ways to set the position of the redgreen edges and C(8,4) ways to set the position of the whitegreen edges, thus forcing the position of the whitered edges. There are 2^11 ways to flip the edges, so we have 495*70*2048 = 70,963,200 Tcosets. The number of essentially different Tcosets reduced by the 48 Oh symmetries can be shown to be 1,482,170.
It is important to observe that each coset of K splits into 70 cosets of T. If you take a look at the pattern example of a coset from K in the table above, there are 8 white edge facelets with the other edge facelet still grey. So you have C(8,4) ways to set the 4 missing red edge facelets and you have no choice left for the missing 4 green edge facelets.

The reduced set cover problem can be formulated in the following way:
Find a subset S of the 64430 essentially different Kcosets, each of them containing 70 Tcosets, such that all all 1,482,170 essentially different Tcosets are covered, in the sense that for each Tcoset t there exists a symmetry m of Oh and a Kcoset k of the subset S such that the coset m' * t * m (which is equivalent up to symmetry to t) is contained in k.
Each of the Kcosets splits again into 2187 Hcosets, and almost all of these Hcosets have to be solved to cover the whole cube space. Why not all 2187 Hcosets in all cases? This happens if the coset K has a D4h symmetry itself. Take for example the identity Kcoset, which is the just the subgroup K itself. It has 16 symmetries, so there are up to 16 different Hcosets which are equivalent to each other.
All these Hcosets are in the same equivalence class for example. 