ag.algebraic geometry - Rational Hilbert modular surfaces

The answer is (probably) yes.

A theorem of Margulis et al. shows that an irreducible lattice in a Lie group is arithmetic unless the group is isogenous to SO(1,n)x(compact) or SU(1,n)x(compact).

A theorem of Mumford shows that the quotient of a hermitian symmetric domain by a neat arithmetic subgroup is of logarithmic general type in the sense of Iitaka (hence not rational). (An arithmetic subgroup is neat if the subgroup generated by the eigenvalues of any element of the subgroup is torsion-free. Every sufficiently small subgroup is neat).

For a discussion of the first theorem, see Section 5B of Witte Morris, Introduction to Arithmetic Groups,
http://front.math.ucdavis.edu/0106.5063

For the second theorem, see: Mumford, D. Hirzebruch's proportionality theorem in the noncompact case. Invent. Math. 42 (1977), 239--272. MR0471627

Added: As @moonface points out, this doesn't prove it.

The congruence subgroup problem is known for $SL_{2}$ over totally real fields $F$ other than $Q$. Let $Gamma$ be an irreducible lattice in $SL_{2}(mathbb{R})times SL_{2}(mathbb{R})$. By Margulis, it is arithmetic, hence congruence. Moreover, after conjugating we may suppose that $GammasubsetGamma(1)=SL_{2}(O)$. Now $Gammabackslashmathbb{H}timesmathbb{H}$ covers $Gamma(1)backslash mathbb{H}timesmathbb{H}$, and so if the first is rational, so is the second (Zariski 1958; requires dimension 2). Thus, we have a finite list of the possible totally real quadratic extensions $K$ to work with. For some $N$, $GammasupsetGamma(N)$, and we may suppose that $Gamma(N)$ is neat. Since $Gamma(1)/Gamma(N)$ is finite, for any integer $N$ we get (in sum) only finitely many rational Hilbert modular surfaces of level $N$.

The problem remains to show that, for each totally real quadratic field on our list, every rational Hilbert modular surface is of level $N$ for some fixed $N$. Apart from looking case by case, I don't see how to do this (but it is surely true).