Homomorphisms from higher rank lattices with infinite center to $\mathbb{Z}$
Yes, every homomorphism $\Gamma \to \mathbb{Z}$ is trivial.
We may assume that $G$ is simply connected, thus it decomposes as a product of simple factors. Let's consider two cases:
$G$ has exactly one non-compact simple factor.
$G$ has at least two non-compact simple factors.
In case 1 $G$ has property (T), so also does $\Gamma$ and the result follows. In case 2 the result follows from theorem 0.8 in
Shalom, Yehuda Rigidity of commensurators and irreducible lattices. Invent. Math. 141 (2000), no. 1, 1–54.
Formally, the above theorem applies only for $\Gamma<G$ cocompact, but in fact the proof shows that you need 2-integrability of $\Gamma$ in $G$, which holds by Proposition 7.1 here, see the preceding discussion for the definition.
The above is an edit of an earlier partial answer I gave, based on the answer of Mikael de la Salle. See Mikael's answer and YCor's comments for further details.
This is a follow-up of Uri's answer. My goal is just to confirm that (for any $p$) the $L^p$-integrability of lattices in a connected semisimple Lie group $G$ follows from the $L^p$-integrability of lattices in $G/Z(G)$. The non-trivial ingredient that is needed is that the central extension $G\to G/Z(G)$ is represented by a bounded $2$-cocycle. The argument (which I think I learned from Nicolas Monod) is at least in Proposition 7.1 of my paper with Tim de Laat https://arxiv.org/abs/1401.3611
The fact that this central extension is represented by a bounded $2$-cocycle follows, for simple Lie groups, from the well-known classical work of Guichardet-Wigner, see also the paper Shtern, A. I. Bounded continuous real 2-cocycles on simply connected simple Lie groups and their applications. Russ. J. Math. Phys. 8 (2001), no. 1, 122–133. The case of semisimple Lie groups follows by decomposing into simple parts.