Suppose you have a fixed sequence of 30 moves. If you do that sequence, you'll presumably go through 30 different states of the cube (possibly not if the sequence cancels itself out, but let's be optimistic and suppose it does not).
Do the same sequence again. You'll go through 30 more states, for 60 total. Do it again and you'll have gone through 90 total, etc. Because there are 43 quintillion possible states of the cube, in some cases you'd have to do the sequence of 30 moves 43 quintillion/30 = about 1.4 quintillion times before solving the cube.
So if you had a sequence of 30 moves that you could do over and over about 10^18 times without going back to where you started, you might be able to do what the original poster wanted.
However, it is not possible to have a sequence of any length that you do one quintillion times without repeating. The reason is simple. Once you do the sequence, you've done a permutation of the edges, and a permutation of the corners, and maybe also changed some of their orientations. For example, the sequence might take three edges and send them around in a cycle (edge 1 goes to edge 2, edge 2 to edge 3, and edge 3 to edge 1), and also move five corners around in a cycle (corner 1 to corner 2, ... until corner 5 goes back to corner 1). If that was what your series of moves did, and you repeated it three times, the edges would all be back where they started. The first time would send edge 1 to edge 2's spot. The second time would send it to edge 3's spot. The third time would send it home. Then, if you repeated the sequence two more times, for five total, the edges would be messed up again, but the corners would be in place. If you did the entire sequence 15 times, the edges and corners would all be back in their starting places.
Those edges and corners might also be flipped or rotated, but if so, by doing the sequence 90 times (15 * 3 *2, three for the corner orientations to be fixed and two for the edges), all the pieces would have to be oriented correctly, and the cube would be solved. This sequence of moves would be said to have "order 90".
It is possible to find things with order higher than 90, but not much higher. The highest order you could achieve is 1260, which would come about if you put the corners in a three cycle and a five cycle, and the edges in a seven cycle and two two cycles, and then made sure that some corner and edge orientations got messed up. There are some technical reasons why you cannot build cycles that would have even higher order - this is the best you can actually achieve. All of what I've said above is sure-fire (unless I made a mistake counting the cycles that make the highest-possible order element of the cube group), and can be made mathematically precise if it's not convincing.
So if the sequence has 30 moves, somewhere in the first 1260 times you do it the cube will go back to solved, and you've only covered 30*1260, or about 40,000 positions. So you could only solve 40,000 positions that way. The shortest thing you could possibly remember that would go through every position would be 4.3*10^19/1260, or 3.4*10^16 moves long. Your computer's hard drive could not store this sequence.
This doesn't mean a sequence this short exists that does the job, only that any sequence that goes through all the positions must be at least this long.
Reading through the link some people provided earlier in the thread, I see the author of the page on the Devil's Algorithm did the same thing with a 2x2 cube to give a minimum length of the sequence you'd need there.
So, I'm not sure whether you'd have to remember a sequence of length 43 quintillion or just 34 quadrillion (or possibly longer than 43 quadrillion, since the "Devil's Number" is unknown) to solve the cube with no decision making (other than knowing when to stop), but either way it is not likely to happen.
