Does $V=L[A]$ imply $H_\kappa = L_\kappa [A]$? It is well-known that (cf. Kunen Set Theory (2013), Theorem II.6.23):

$V=L$ implies that $H_\kappa = L_\kappa$ for all cardinals $\kappa \geq \omega$.

I wonder if this can be generalized for relative constructibility $(*)$ in the following sense:

Let $A$ be an arbitrary set. $V=L[A]$ implies that $H_\kappa = L_\kappa [A]$ for all cardinals $\kappa \geq \omega$.

I've tried to generalize the proof of the first theorem, and I think that at this level of generality the second statement is not true (the problem here is that Condensation lemma works a bit different with relative constructibility. See e.g. here). Even assuming that $A \in L[A]$ is probably not sufficient. But maybe if $A$ is a set of ordinals, or even more $A \subseteq \omega_1$, then the proof can be generalized. I'm not sure how to do it, though.
$(*)$ $L[A]$ is defined recursively by $L_0 [A] = \emptyset$, $L_{\alpha+1}[A] = \operatorname{def}_A (L_\alpha [A])$, $L_\eta = \bigcup_{\alpha < \eta} L_\eta [A]$ for $\eta$ limit. By $\operatorname{def}_A (X)$ we mean the family of subsets of $X$ which are definable over $(X,\in, A \cap X)$, where $A \cap X$ is seen as a unary predicate.
 A: No. Consider $L[U]$ as a canonical model with a measurable where $U$ is a measure. In this model you have $0^\#$, which is a subset of $\omega$. 
However, $L_{\omega_1}[U]=L_{\omega_1}$, in which we don't have $0^\#$. 
A: You don't need large cardinals for this, just $V\ne L.$
So assume $V \ne L.$  Then there is a non-constructible set of ordinals (since a transitive model of ZFC is determined uniquely by the sets of ordinals it has).
Let $S$ be a non-constructible set of ordinals, and let $\kappa$ be a cardinal greater than $\cup S.$  Define $A=\lbrace \kappa + \alpha \mid \alpha \in S\rbrace.$
We'll work in $L[A],$ and we'll see that, in $L[A],$ $\,L_\kappa[A] \ne H_\kappa.$
$A\cap\kappa=\emptyset,$ so $L_\kappa[A]=L_\kappa,$ and it follows that $S\not\in L_{\kappa}[A].$
We have $S \in L[A],$ since $S=\lbrace x \in L_{\kappa + \kappa}[A] \mid x\text{ is an ordinal }\lt \kappa\text{ and }\kappa + x \in A\rbrace,$ so that $S$ is first-order definable over $\langle L_{\kappa + \kappa};\, \in;\, A\rangle.$
Since $\kappa$ is a cardinal in the universe, it's a cardinal in $L[A],$ so $\cup S\lt\kappa$ implies immediately that, in $L[A],$ $S$ is hereditarily of cardinality less than $\kappa,$ that is, $S\in H_\kappa.$

If $A\subseteq \omega_1,$ things are different. In the argument above, if we start with a non-constructible real, we get a counterexample $A\subseteq\omega_1+\omega.$
However, there is no counterexample with $A\subseteq \omega_1.$ In fact, if $V=L[A]$ where, for some $n<\omega,$ $A\subseteq\omega_1+n,$ then for every infinite cardinal $\kappa,$ $L_\kappa[A]=H_\kappa,$ by the following argument.
First, we can assume that $n=0,$ so that $A\subseteq\omega_1,$ since adding finitely many additional ordinals to $A$ doesn't change either  $L[A]$ or any $L_\kappa[A].$
Assume $V=L[A],$ and let $\kappa$ be an infinite cardinal.  Clearly $L_\kappa[A]\subseteq H_\kappa.$
To show that $H_\kappa \subseteq L_\kappa[A],$ let $X$ be some member of $H_\kappa,$ and we'll show that $X$ belongs to $L_\kappa[A].$
If $\kappa = \omega,$ this is easy.
If $\kappa \ge \omega_2,$ essentially the same argument as for $L$ works: Let $M$ be the Skolem hull of $\omega_1 \cup \operatorname{TC}(\lbrace X\rbrace)$ in $\langle L_\kappa[A];\,\in;\,A\rangle,$ where $\operatorname{TC}(\lbrace X\rbrace)$ denotes the transitive closure of $\lbrace X\rbrace.$  $M$ has cardinality less than $\kappa,$ so its Mostowski collapse must be some $L_\gamma[B]$ where $\gamma\lt\kappa.$  Every countable ordinal is fixed by the collapse isomorphism in this case, and $A\subseteq\omega_1,$ so $B=A$ by elementarity. Finally, the Mostowski collapse is the identity on $\operatorname{TC}(\lbrace X\rbrace),$ by an $\in\!\text{-induction},$ so $X\in L_\gamma[B]=L_\gamma[A]\subseteq L_\kappa[A].$
For $\kappa=\omega_1,$ however, this argument needs to be modified, since $B$ may not equal $A$ any more.  Instead of just taking the Skolem hull to get an elementary submodel, we'll build an elementary chain $M_0\prec M_1 \prec M_2 \prec \dots$ of length $\omega,$ with each $M_j$ being a countable elementary submodel of  $\langle L_{\omega_1}[A];\,\in;\,A\rangle.$  Let $M_0$ be the Skolem hull of $\operatorname{TC}(\lbrace X\rbrace)$ in $\langle L_{\omega_1}[A];\,\in;\,A\rangle.$ To define $M_{j+1},$ let $\eta_j$ be the countable ordinal $\bigcup(M_j\cap\omega_1),$ and let $M_{j+1}$ be the Skolem hull of $\eta_j \cup \operatorname{TC}(\lbrace X\rbrace) \cup M_j$ in $\langle L_{\omega_1}[A];\,\in;\,A\rangle.$  (We don't really need to include $M_j$ in the union here, since it will be contained in the Skolem hull anyway, but doing it this way avoids having to argue that $M_j\subseteq M_{j+1}.)$  Let $M=\cup_{j\lt\omega}M_j,$ and write $\eta=\cup_{j\lt\omega}\eta_j.$   By the construction, $\langle M;\,\in;\,A\cap M\rangle \prec \langle L_{\omega_1}[A];\,\in;\,A\rangle,$ and $M\cap\omega_1=\eta;$ the Mostowski collapse of $M$ is $L_\eta[B]$ for some $B.$ The collapse isomorphism here is the identity on $\eta,$ and it follows by elementarity that the Mostowski collapse is $L_\eta[A\cap\eta];$ the rest of the argument works the same way as for $\kappa\ge\omega_2.$
