The rotation of dice on a grid

Here's a slick method: label the die with the usual numbers of pips (so 1 is opposite 6, 2 is opposite 5, and 3 is opposite 4) and label the grid like a chessboard. Then place the die on a black square with the die angled in an isometric view (think Q*bert) with the numbers 1,4,5 visible (for instance).

Now, what happens if you make a 90 rotation around an edge? Two of the faces remain visible, but the face which is hidden is replaced with its opposite face. The important point to note: opposite faces have opposite parity (odd/even), so the resulting sum of visible faces now also has opposite parity!

(In the above image, note that the 4(even) is replaced with a 3(odd), changing the sum of the visible faces from 10(even) to 9(odd).)

Thus, no matter how you manipulate the die, it will always have an even sum of visible faces when positioned on the black squares, and it will always have an odd sum of visible faces when positioned on the white squares. Of course, if the die began instead with an odd sum of visible faces on a black square or an even sum of visible faces on a white square, then it will always have an odd sum of visible faces when positioned on the black squares and an even sum of visible faces when positioned on the white squares.

That is sufficient to answer your question. It's not hard to show further that there exist round trips to rotate a die into any configuration of the same parity.

(As an aside, this is a useful trick to solving a puzzle from Zero Escape: Virtue's Last Reward.) [1] [2]

Rolling around a square produces a $\frac 13$ rotation about a body diagonal. This generates a subgroup of size three. Rolling around a different square will fix a different diagonal, generating another order three subgroup. Both of these are subgroups of the even permuations. I believe they generate the whole even subgroup. Rolling around a $2 \times 1$ rectangle produces a $\frac 12$ rotation around an axis through opposite face centers. This generates a subgroup of size six. Both of these correspond to even permutations, while a single quarter turn is an odd permutation. By checkerboard coloring the grid, we know that to return to the starting square requires an even number of quarter turns, so a single quarter turn is not possible.

Added:It is actually much easier for other side dice. For all of the other Platonic solids, you have specified the orientation once you specify the bottom face. For $20$ side dice, if you roll six times around one vertex, you come back to start having moved one face, so you can get any face on the bottom at the starting point. For $4$ side dice, when you come back to start you have the same face down as you started with. You can cover the plane with a triangular grid in the standard way and each triangle has a face that will always be down when the tetrahedron is on that cell. For $12$ side dice I don't think you can get back to the starting location except by retracing your steps from somewhere because the pentagons don't overlay. For $8$ sided dice you can color alternate faces black and white on the octahedron and only the faces that match the original one in color can wind up down at the starting point-color the triangular grid on the plane to match. As rolling six times around the same point gets you back to start with a different face down, you can get four faces down at start.