The claim you are trying to prove is false: $L(\mathbb R)$ has no measurable cardinals greater than $\Theta$. Work in $L(\mathbb R).$ We will use Woodin's theorem that $\text{HOD} = L[\mathbb P]$ for $\mathbb P\subseteq \Theta$ a partial order encoded as a subset of $\Theta$ and that $L(\mathbb R)$ is an inner model of $\text{HOD}[G]$ for a $\text{HOD}$-generic filter $G\subseteq \mathbb P$. (We will use the second half of this at the end of our answer but for now it suffices to accept that $\text{HOD} =L[A]$ for $A\subseteq \Theta$.) If $\kappa>\Theta$ is measurable, and $U$ is a $\kappa$-complete ultrafilter on $\kappa$ then we have $\text{Ult}(\text{HOD},U) = L[j(\mathbb P)] = L[\mathbb P] = \text{HOD}$ where $j$ is the ultrapower map. Now $j:\text{HOD}\to \text{HOD}$ witnesses $\mathbb P^\#$ exists. But then $\mathbb P^\#$ is in $\text{HOD}$, contradicting $\text{HOD} = L[\mathbb P]$.

The issue with your proof is that in general $L(\mathbb R)[G]$ is a proper subset of $L(\mathbb R)^{V[G]}$ when $G$ is a $V$-generic real. (There is no reason that names for reals should be reals themselves.) It fails in particular for the situation in interest: $G\subseteq \text{Col}(\omega,\kappa)$ (assuming a proper class of Woodin cardinals in $V$). At least, if $\kappa \geq |\mathbb R|$, then in $V[G]$ there is a real $r$ coding an enumeration of $\mathbb R^V$ in order type $\omega$, but if there is such a real in $L(\mathbb R)[G]$, then $L(\mathbb R)[G]$ satisfies AC (since then $L(\mathbb R)[G] = L[r,G]$). Thus assuming $L(\mathbb R)[G] = L(\mathbb R)^{V[G]}$ we obtain $L(\mathbb R)^{V[G]}\vDash \text{AC}$, a contradiction in the context of our large cardinal hypothesis. (If $V = L$ we of course have no contradiction at all.)

Incidentally, if $L(\mathbb R)\vDash \text{AD}$ then $\Theta$ is not measurable either. Suppose it were. Consider $j:\text{HOD}\to \text{Ult}(\text{HOD},U)$ where $U\in L(\mathbb R)$ is a $\Theta$-complete ultrafilter on $\Theta$. Since $\Theta$ is inaccessible in $\text{HOD}$, $j(\Theta)$ is inaccessible in $\text{Ult}(\text{HOD},U)$ and hence in $\text{HOD}$ since $\text{HOD} = L[\mathbb P] \subseteq L[j(\mathbb P)] = \text{Ult}(\text{HOD},U)$. On the other hand $j(\Theta)$ is not a cardinal in $L(\mathbb R)$ since $|j(\Theta)| = | P(\Theta)\cap \text{HOD}| = |(2^\Theta)^{\text{HOD}}|$, while $(2^{\Theta})^{\text{HOD}} < j(\Theta)$ (as ordinals) again since $j(\Theta)$ is inaccessible in $\text{HOD}$. Now $L(\mathbb R)$ is an inner model of $\text{HOD}[G]$ for a $\text{HOD}$-generic filter $G\subseteq \mathbb P$. In $\text{HOD}[G]$, $j(\Theta)$ remains inaccessible by the usual Levy-Solovay analysis. This contradicts the fact that $j(\Theta)$ is not a cardinal in the inner model $L(\mathbb R)\subseteq \text{HOD}[G]$.

The claim you are trying to prove is false: $L(\mathbb R)$ has no measurable cardinals greater than $\Theta$. Work in $L(\mathbb R).$ We will use Woodin's theorem that $\text{HOD} = L[\mathbb P]$ for $\mathbb P\subseteq \Theta$ a partial order encoded as a subset of $\Theta$ and that $L(\mathbb R)$ is an inner model of $\text{HOD}[G]$ for a $\text{HOD}$-generic filter $G\subseteq \mathbb P$. (We will use the second half of this at the end of our answer but for now it suffices to accept that $\text{HOD} =L[A]$ for $A\subseteq \Theta$.) If $\kappa>\Theta$ is measurable, and $U$ is a $\kappa$-complete ultrafilter on $\kappa$ then we have $\text{Ult}(\text{HOD},U) = L[j(\mathbb P)] = L[\mathbb P] = \text{HOD}$ where $j$ is the ultrapower map. Now $j:\text{HOD}\to \text{HOD}$ witnesses $\mathbb P^\#$ exists. But then $\mathbb P^\#$ is in $\text{HOD}$, contradicting $\text{HOD} = L[\mathbb P]$.

The issue with your proof is that in general $L(\mathbb R)[G]$ is not equal to $L(\mathbb R)^{V[G]}$ even when $G$ is a $V$-generic real. (There is no reason that names for reals should be reals themselves. Furthermore there is no reason that $\mathbb R^V$ should be in $L(\mathbb R)^{V[G]}$.) It fails in particular when $G\subseteq \text{Col}(\omega,\kappa)$ (assuming a proper class of Woodin cardinals in $V$). At least, if $\kappa \geq |\mathbb R|$, then in $V[G]$ there is a real $r$ coding an enumeration of $\mathbb R^V$ in order type $\omega$, but if there is such a real in $L(\mathbb R)[G]$, then $L(\mathbb R)[G]$ satisfies AC (since then $L(\mathbb R)[G] = L[r,G]$). Thus assuming $L(\mathbb R)[G] = L(\mathbb R)^{V[G]}$ we obtain $L(\mathbb R)^{V[G]}\vDash \text{AC}$, a contradiction in the context of our large cardinal hypothesis. (If $V = L$ we of course have no contradiction at all.)

Incidentally, if $L(\mathbb R)\vDash \text{AD}$ then $\Theta$ is not measurable either. Suppose it were. Consider $j:\text{HOD}\to \text{Ult}(\text{HOD},U)$ where $U\in L(\mathbb R)$ is a $\Theta$-complete ultrafilter on $\Theta$. Since $\Theta$ is inaccessible in $\text{HOD}$, $j(\Theta)$ is inaccessible in $\text{Ult}(\text{HOD},U)$ and hence in $\text{HOD}$ since $\text{HOD} = L[\mathbb P] \subseteq L[j(\mathbb P)] = \text{Ult}(\text{HOD},U)$. On the other hand $j(\Theta)$ is not a cardinal in $L(\mathbb R)$ since $|j(\Theta)| = | P(\Theta)\cap \text{HOD}| = |(2^\Theta)^{\text{HOD}}|$, while $(2^{\Theta})^{\text{HOD}} < j(\Theta)$ (as ordinals) again since $j(\Theta)$ is inaccessible in $\text{HOD}$. Now $L(\mathbb R)$ is an inner model of $\text{HOD}[G]$ for a $\text{HOD}$-generic filter $G\subseteq \mathbb P$. In $\text{HOD}[G]$, $j(\Theta)$ remains inaccessible by the usual Levy-Solovay analysis. This contradicts the fact that $j(\Theta)$ is not a cardinal in the inner model $L(\mathbb R)\subseteq \text{HOD}[G]$.