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][1]

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.


  [1]: http://www.mathematik.tu-darmstadt.de/fbereiche/AlgGeoFA/staff/stroemberg/research/gallery/