Maass forms will generally have nodal lines rather than zeros. Let's take the case where $\Gamma = SL_2(\mathbb{Z})$ and the weight $k=0$.
First recall that a Maass form has a Fourier expansion of the form $$f(x+iy) = \sum_{n \neq 0} a_n \sqrt{y} K_{\nu}(2 \pi |n| y) e^{2 \pi i n x},$$ where $K_{\nu}$ is the usual $K$-Bessel function. Here $\nu$ is purely imaginary. It is not difficult to show that $K_{it}(v)$ is real for $t, v$ real. If $f$ is an eigenfunction of all the Hecke operators then the $a_n$'s are real also, which is a consequence of the fact that the Hecke operators are self-adjoint with respect to the Petersson inner-product. Furthermore, by considering the reflection operator $T_{-1}f(x+iy) = f(-x+iy)$ which commutes with all the Hecke operators as well as the Laplacian, we should suppose $f$ is an eigenfunction of $T_{-1}$ also. The only possible eigenvalues of $T_{-1}$ are $\pm 1$ so in the above Fourier expansion we can replace $e^{2 \pi i n x}$ by $\cos(2 \pi n x)$ or $\sin(2 \pi n x)$. Then in either case $f$ is patently real-valued.
One can find some nice pictures of Maass forms at Fredrik Strömberg's homepage
Edit: I should also point out the simple fact that if $f$ is odd (i.e. $f(-x+iy) = -f(x+iy)$) then $f(iy) = 0$ for all $y$. This can be clearly seen in Strömberg's pictures.