I thought you wanted a set of cubes whose volumes don't intersect with each other. (Then this set can't be uncountable because each volume contains a rational point.) I don't understand what you mean by replacing cubes... Can you explain the problem once again?
Assumed, that the ZF is consistent, you are right, of course. The fact, that there are rational points inside every cube, and that those cubes have no common volume, only common surfaces (lines, points) perhaps - is enough to state that there are at most aleph-zero of them.
But, if the ZF is not consistent, there may be a way to encode every real number between 0 and 1 with some peculiar cubic subdivision of 3D space.
You can easily encode every real between 0 and 1 via dividing (even a finite volume of) 3D space into disjunct 2D circles.
Perhaps, just perhaps, it is possible to replace those circles with cubes and spheres, given the infinite volume you have.
Perhaps, just perhaps, the ZF is broken.
If it's worth saying, but not worth its own post, then it goes here.
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