I don't think it's possible to find a "nice" (say, smooth) function $f \in L_2(\Gamma \backslash \mathbb{H})$ such that $(1) \int_0^{1} f(x+iy) dx = 0$ for all $y > 0$ and $\lim_{y\rightarrow \infty} f(x+iy) \neq 0$.  This may be total overkill, but consider the spectral decomposition of such an $f$, namely
$$(2) \qquad f(z) = \sum_{j} \langle f, u_j \rangle u_j(z) + \frac{1}{4\pi } \int_{\mathbb{R}} \langle E(\cdot, 1/2 + it), f\rangle E(z, 1/2 + it) dt.$$
By unfolding, the inner product of $f$ with the Eisenstein series $E(z,s)$ is zero by the assumption (1); initially this is easy for the real part of $s$ large but then follows by analytic continuation.
By inserting (2) into (1) we see that $\langle f, u_0 \rangle = 0$, that is $f$ is orthogonal to the constant eigenfunction.  Now in (2) take $z= x+iy$ with $y$ large.  Each term in the sum is very small since all the Maass forms vanish at the cusp, and the projections of $f$ onto the constant eigenfunction and the Eisenstein series are zero.