Question
Suppose that $\Gamma < \text{SU}(n,1)$ is a cocompact lattice, and let $\rho \colon \Gamma \to G$ be a representation to a non-compact simple Lie group (most interesting case for me: $G = \text{SU}(p,p)$). Suppose that $\rho$ is Zariski-dense. Is it true that $\rho$ is not obstructed, that is, every infinitesimal deformation in $H^1(\Gamma, \text{Ad}(\rho))$ is tangent to an actual family $\rho_t \colon \Gamma \to G$ with $\rho_0 = \rho$?
Background and motivation
The question can be posed more generally for Kähler groups or for more general (cocompact) lattices $\Gamma$, but the hyperbolicity hypothesis could help. By Goldman-Millson theory, this would mean that the bracket map
$$
H^1(\Gamma, \text{Ad}(\rho)) \times H^1(\Gamma, \text{Ad}(\rho)) \to H^2(\Gamma, \text{Ad}(\rho))
$$
vanishes identically, but I think there is no hope that $H^2$ vanishes itself (quite to the contrary, Carlson-Toledo conjecture expects $H^2(\Gamma, \mathbb{R}) \neq 0$ always, but this does not say much about the above map).
I ask this because the only two examples of rigid but not infinitesimally rigid representations I know of are the one constructed by Goldman and Millson, who considered the embedding $\text{SU}(n,1) \hookrightarrow \text{SU}(n+1,1)$ and the one constructed by Kim, Klingler and Pansu, that is $$ \text{SU}(n,1) \hookrightarrow \text{Sp}(n,1) \hookrightarrow \text{SU}(2n,2). $$ However, in both cases the representation comes from an infinitesimally rigid representation (in the first case: up to "trivial deformations") to a smaller subgroup of $\text{SU}(p,q)$. The $H^1$ in this bigger group turns out to be non-zero, and along such directions there are obstructions, but this does not happen if we restrict to the Zariski closures. Furthermore, in both cases the representation factors through the whole of $\text{SU}(n,1)$, an hypothesis I am not assuming.
