EDIT: This only works in a certain set of configurations, as the loop only goes through 1260 states before returning to the original position. My mistake, this is incorrect, but still useful.
Great solution found here. Basically, if you rotate the right, back, left, and front faces all clockwise and in order, then it will always solve the cube - eventually. The core part of it is in one simple explanation - doing this cannot enter an inescapable loop.
If you move the front face clockwise, then counter-clockwise, nothing happens to the cube. The faces are not moved with respect to each other, and you are in the same position. In order for a loop to happen, two different layouts must lead to the same configuration with the same move. But wait - if you reverse this move, then what position does it go to? The simple answer is this will never happen. Here is a great diagram in the article itself. 
As you can see, reversing this loop would not work, as there are two possible options. This is a proof that loops will never happen.
Moving these four faces is the shortest possible configuration of moves that affects each and every cube, save for the center ones which are irrelevant anyway. This means it is the shortest possible solution that you could repeat to solve the Cube.
<20 move sequence>
followed by<undoing previous sequence>
) as a possible solution? (might not be shortest but it'll show one exists) You'll solve the cube at some point. $\endgroup$