No, it's sufficient to consider 3 "rotations" of each box, because the only possibilities to consider are which dimension to make the top and bottom (say) of the box perpendicular to, and there are only 3 of these. It might help to think of the 3 different possibilities for each box in that way, instead of as rotations.
The main point is that, after we have chosen which pair of sides of a box will be top and bottom (which we can do in 3 ways), we don't need to try the 2 different rotations of the box in the plane. We can always just go with (say) the rotation in which the box is wider than it is deep. How do we know that we won't miss any potentially good solutions by doing this? Because in any solution having a box that is deeper than it is wide, there must be a lowest such box b, and that box, plus everything above it, can be safely rotated 90 degrees. (We know it's safe to do this because it's the lowest such box in the solution -- so the box underneath it, if any, must itself be wider than it is deep, meaning that if we rotate b 90 degrees, it must still fit inside this lower box (I recommend verifying this algebraically).) We can keep transforming the lowest deeper-than-it-is-wide box in the solution until none remain, without ever changing the height of the solution. Since we can do this for any solution containing one or more deeper-than-it-is-wide boxes, it means that each one is exactly equivalent in quality to some solution in which every box is wider-than-it-is-deep, so we can totally ignore the former solutions.