In the subtraction problem, we can start by looking at the left column. The first possible case is that $ A-B=0 $, so that $ A=B $. However, if $ A=B $, that does not work in the right column, since $ B-A $ would equal 0 and not 3. So, we have to go with the second possibility for the left column, in which the right column borrowed from the left column and we're left with $ (A-1)-B=0 $. This tells us that $ A-1=B $, so the nonnegative difference of $ A $ and $ B $ is $ \boxed{1} $. Note that we are unable to solve for the actual digits $ A $ and $ B $ even if we consider the units digit. Now that we know the right column borrows from the left column, we get that $$ \begin{array}{c@{}c@{\;}c@{}c@{}c@{}c} & & & & 1 & B_4\\ &- & & & & A_4\\ \hline & & & & & 3_4.\\ \end{array} $$This also means $ A+3=1B_4=4+B $, which is the same as $ A-1=B $. We can't solve for $ A $ and $ B $ because any digits that satisfy $ A-1=B $ will work, such as $ A=3, B=2 $ or $ A=2, B=1 $.