Take 6 squares of identical size and label each of them with a different non-symmetric mark, for example: $1,2,3,4,5,6$ (writing $3$ in a way such that is posseses no internal non trivial symmetry). How many pair-wise non-congruent cubes, including the marks, can be built from these 6 squares?
The question is inspired by a puzzle game consisting of 6 pieces that should be put together to a cube: 
My idea would be that for each square you have $$ 4\cdot 2$$ possibilities to orient it (including the rotation between front side and back side). And then there are 6! possibilities to arrange each of them. But then there are 24 symmetry equivalent possibilities for each cube (this is the order of the point group $O$). Hence $$ \frac{(4\cdot2)^6 6!}{24} = 7864320\approx 8\cdot10^6,$$ which seems intuitively (by far) too many.
How to best tackle this?
Edit: Thinking about it for a bit longer, the most concise definition of the problem is probably "tiling a cube with non-identical non-symmetric tiles"