$(FC)_{\text{proper}}$ is false. If there are infinitely many Woodin cardinals with a measurable above, then after adding a single Cohen real, we do not have $L(\mathbb R)[c] = L(\mathbb R)^{V[c]}$. The reason is that $L(\mathbb R)[c]$ does not satisfy $\text{AD}$: in $L(\mathbb R)[c]$, the set of ground model reals is uncountable, but does not contain a perfect subset. (See Hamkins's answer to the question https://mathoverflow.net/questions/251988/is-there-a-perfect-set-of-ground-model-reals-in-the-cohen-extension which uses at worst countable choice.) On the other hand, $L(\mathbb R)^{V[c]}$ satisfies $\text{AD}$ by our large cardinal hypothesis. 

Note in this situation that $L(\mathbb R)^{V[c]}=L(\mathbb R)^{L(\mathbb R)[c]}\subsetneq L(\mathbb R)[c]$, since nice names for reals are all coded by reals. Therefore it is *not* true in general that $L(\mathbb R)^{L(\mathbb R)[r]} = L(\mathbb R)[r]$ (or that $L(\mathbb R)[r]\subseteq L(\mathbb R)^{V[r]}$) when $r$ is a generic real. The problem this time is that $\mathbb R$ (as computed in the ground model) need not be in $L(\mathbb R)^{L(\mathbb R)[r]}$.

The large cardinal assumption can actually be reduced to the assumption that $\text{AD}$ holds in $L(\mathbb R)$, because it can be proved under this hypothesis that $ L(\mathbb R)^{V[c]} = L(\mathbb R)^{L(\mathbb R)[c]}$ satisfies $\text{AD}$. You might be interested in Kechris and Woodin's paper Generic Codes, in which they prove this fact and analyze forcing over $L(\mathbb R)$ with Levy collapses, showing that for $\kappa<\delta^2_1$ a reliable ordinal, if $G\subseteq \text{Col}(\omega, \kappa)$ is $L(\mathbb R)$-generic then in $L(\mathbb R)[G]$, there is a definable elementary embedding from $L(\mathbb R)$ into $L(\mathbb R)^{L(\mathbb R)[G]}$