What is the probablity that a Rubik's cube is solvable if you randomly switch two squares? I am trying to learn how to solve a Rubik's cube and the 4x4 rubic cube at school is missing 4 squares, that got me thinking: if you randomly switched two squares, would it still be solvable?  
I know that the flat picture (example of what I mean) where you slide a square around on the xy plane are solvable omitting if you switch the last two squares when it is in its identity location. For example if cut a picture into 16 squares and then removed the 16th (bottom righthand) and then you switched the 14th and the 15th square then it would be unsolvable. 
I really don't know how to say that better mathematicaly 

I'm posting a new answer, because existing answer deals with 3×3 cube, which is very different in this regard from a 4×4 cube.
To actually answer the question, YES a 4×4 cube will be solvable if you randomly exchange any two centers. 
On a 4×4 cube, the parity of centerpieces permutation must always be the same as the parity of the cornerpieces permutation, however, odd and even parity of centers permutation is not distinguishable, as there are 4 identical centers of each color.
This can be demonstrated by a simple commutator-algorithm such as this one. This algorithm exchanges three centerpieces from the mechanical perspective, however, due to the fact that two of those are of the same color, the resulting effect is that only two centerpieces have been exchanged.
