Turing machine that accepts L = {#a != #b}
You dont need to keep track of them. At each iteration you have three possibilities. Either you first see an A, or a B or a nothing. If nothing accept. If an A first, replace it with some special character and search for a B. If you see one replace that one as well with some special character go back to beginning but if you dont just reject. Do the same for B first. After you cross our 2 characters loop back to beginning. It is actually very similar to $a^nb^n$.

The TM design should be straightforward actually. Similar to $a^nb^n$ but considers $b$ first option as well at each iteration.