Is writing a number as two squares and writing the factors of a number equally hard?
Here's the situation as far as I can tell:

The most efficient way known to test membership in $L_1$ is to factor $r$.  No more efficient algorithm seems to be known.

However, there is no obvious reduction to prove $L_2 \le_P L_1$.  In other words, if we had a machine to decide $L_1$, there's no easy way to use that to factorize.  If we want to factor a number $N$, we can check whether $N \in L_1$, but this only gives us one bit of information about the factors of $N$.  Testing multiples of $N$ (i.e., whether $cN \in L_1$ for some integer $c$) doesn't give us any additional information, as knowing whether $N \in L_1$ suffices to predict whether $cN \in L_1$ for all $c&gt;0$.  Testing other numbers doesn't seem to help in any obvious way, either. (Testing whether $N' \in L_1$ for some other number $N'$ doesn't seem to give any useful information, if $\gcd(N,N') =1$; and if we had a way to find another number $N'$ such that $\gcd(N,N')\ne 1$, we'd already know how to factor, but of course we don't know how to do that -- so any number we know how to generate is unlikely to give us any useful information on the factors of $N$.)  This is not a proof; it is just handwavy intuition.

So it seems plausible that $L_1$ might be easier than $L_2$, and it also seems plausible that they might be of the same difficulty.  I'm not aware of any further results on this subject.