It is known that $f(n):=r_2(n)/4$ is a multiplicative function such that for $p\equiv 1\pmod{4}$ we have $f(p^k)=k+1$, while for $p\equiv 3\pmod{4}$ we have $f(p^k)=1$ or $f(p^k)=0$ depending on whether $k$ is even or odd. Using this information, one can show that
$$r_2(n)\leq n^{\frac{\log 2+o(1)}{\log\log n}},$$
and this is best possible in the sense that $\log 2$ cannot be lowered here. The proof goes almost verbatim as the proof of Theorem 2 in Section 5.2 in Tenenbaum: Introduction to analytic and probabilistic number theory. In fact the statement of this theorem itself implies the above upper bound, because $f(n)\leq\tau(n)$. The sharpness of $\log 2$ only requires a Chebyshev type lower bound that
$$\sum_{\substack{p\leq x\\p\equiv 1\pmod{4}}}\log p\gg x.$$

Regarding Noam Elkies's comment: Landau proved that the number of $n\leq x$ with $r_2(n)>0$ is asymptotically
$$2^{-1/2}\prod_{p\equiv 3\pmod{4}}(1-p^{-2})^{-1/2}\frac{x}{\sqrt{\log x}}.$$
For a proof, see Section 1.8 in Brüdern: Einführung in die analytische Zahlentheorie.

GH from MOsupplied a more thorough answer with sources, so I refer you to his answer for further information. $\endgroup$ – Noam D. Elkies Dec 23 '14 at 17:00