Can you do an in-place reversal of a string on a vanilla turing machine in time $o(n^2)$?
The standard way to prove things like this is to show that $\Theta(n)$ bits of information must cross at least $\Theta(n)$ points. 

That is, if you're reversing it "in place", the first third of the bits must cross all $n/3$ middle cells and end up in the last $n/3$ cells. Since the head moves $O(1)$ bits a distance at most $1$ cell each step, this requires $n^2/9$ steps. If the reversed copy ends up in a different set of cells, a very similar argument gives an $\Omega(n^2)$ lower bound.

It takes some work to make this intuitive idea rigorous, but it has been done, although I don't have enough time to locate the papers these results appeared in. If anybody can point to a relevant paper, please edit this answer to do so.