This question seems to be a little more subtle than the usual homogeneity arguments. Here is one way that I found to do it.$\newcommand\R{{\mathbb{R}}}\newcommand\HOD{\text{HOD}}\newcommand\Ord{\text{Ord}}$
Allow me to describe the problem like this. We start in $L$ and force with
$\newcommand\P{\mathbb{P}}\P=\text{Add}(\omega,\kappa)$ to add
$\kappa$ many Cohen reals. In the forcing extension $L[G]$, we form
the class $\HOD_\R^{L[G]}$ of hereditarily ordinal-definable sets.
Meanwhile, $L(\R)$ is the class of sets constructible from reals in $L[G]$,
defined by starting with the real numbers of $L[G]$ and iteratively taking
the definable power set.
Theorem. $\HOD_\R^{L[G]}=L(\R)$.
Proof. The inclusion $L(\R)\subseteq\HOD^\R$ is easy to see, as
alluded to in the question, because every real is definable using
itself as a parameter and, as can be seen by induction, every
object in $L(\R)$ is definable in $L(\R)$ using finitely many
ordinal and real parameters, and so every object there is definable
in $L[G]$ using real and ordinal parameters. Indeed, in $L(\R)$
there is a definable surjection $s:\Ord\times\R\to L(\R)$, and this
surjection is definable in $L[G]$.
It remains to see, conversely, that $\HOD^\R\subseteq L(\R)$.
Suppose that $A$ is hereditarily definable in $L[G]$ using ordinal
and real parameters. By $\in$-induction, we may assume that
$A\subseteq L(\R)$. Let $B=\{(\alpha,z)\mid s(\alpha,z)\in A\}$,
which is also definable in $L[G]$, using the definable surjection.
It suffices to show that $B\in L(\R)$.
Since $B\in\HOD_\R^{L[G]}$, there is a formula $\varphi$ for which
$$B=\{(\alpha,z)\mid L[G]\models\varphi(\alpha,z,\beta,a)\},$$
for some ordinal parameter $\beta$ and real parameter $a$. Fix
names $\dot B$ and $\dot a$ for which $\dot B_G=B$ and $\dot
a_G=a$, and fix a condition $p_0\in G$ forcing that $\dot B$ is
defined by $\varphi(\cdot,\cdot,\check\beta,\dot a)$.
Claim. $(\alpha,z)\in B$ if and only if there is $\dot z$ and
$L$-generic $g$ for a countable complete suborder of $\P$ in $L$, with
$p_0\in g$ and $\dot a_g=a$ and $\dot z_g=z$ and $\exists p\in g$
with $L\models p\Vdash \varphi(\check\alpha,\dot z,\check\beta,\dot a)$.
Proof. The forward implication is easy, since we can take $g$
to be a fragment of $G$. Conversely, suppose that we have $g$ and
$\dot z$ as stated. Since $g$ is an $L$-generic real in $L[G]$, we
may extend $g$ to a full $L$-generic filter $H\subset\P$ with
$L[G]=L[H]$. It follows that there is an automorphism of the
forcing $\pi:\P\cong\P$ in $L$ with $\pi[H]=G$. Since
$p\Vdash\varphi(\check\alpha,\dot z,\check\beta,\dot a)$, it
follows that $\pi(p)$ forces $\varphi(\check\alpha,\dot
z^\pi,\check\beta,\dot a^\pi)$, where the superscript $\pi$ means
that we have applied the induced automorphism on names.
Since $\dot a_G=a=\dot a_g=\dot a_H=\dot a^\pi_{\pi[H]}=\dot
a^\pi_G$, it follows that there is some condition $q\in G$ forcing
that $\dot a=\dot a^\pi$. Without loss, $q\leq \pi(p)$. So $q$
forces $\varphi(\check\alpha,\dot z^\pi,\check\beta,\dot a^\pi)$.
Since $z=\dot z_g=\dot z_H=\dot z^\pi_{\pi[H]}=\dot z^\pi_G$, this
implies $(\alpha,z)\in B$, as desired. QED (claim)
The claim implies the theorem, since that definition can be carried
out in $L(\R)$, using $a$ and $\beta$ as parameters and the fact that the forcing relation is definable in $L$. QED
This argument is a little more complicated than I expected, and I'm not sure if there is a more direct argument.
