It is a fantastic question, Jonas! I've spent hours with it now, going back and forth several times about which way it might go.
But finally, I've got a negative answer, at least for some models $M$. My idea is that some models of ZFC admit what are called satisfaction classes, but these can never be added by class forcing.
For any transitive model $M$ of ZFC, let $S$ be the set of pairs $\langle{}^\ulcorner\varphi{}^\urcorner,a\rangle$, for which $\varphi[a]$ holds in $M$. This class $S$ is the (unique) class of pairs $\langle n,a\rangle$ satisfying the Tarskian inductive definition of truth. Further, we may sometimes add $S$ as a class to $M$ and still have GBC in $(M,S)$. This is true, for example, when $M=V_\kappa$ for an inaccessible cardinal $\kappa$; but also it is true for many other models.
So let us suppose that $M$ admits such a unique satisfaction class.
Can $S$ be added by class forcing over $M$ without adding sets? If we regard $M$ as a GB model by endowing it with only its definable classes, then the answer is no.
To see this, suppose that $S$ was added in the class forcing extension $M[H]$, where $H\subset\mathbb{P}$ is $M$-generic for the class forcing $\mathbb{P}$, which adds no sets. So $S=\dot{A}_H$ for some class $\mathbb{P}$-name $\dot{A}$, and in particular, both $\dot{A}$ and $\mathbb{P}$ are definable in $M$.
Now, the key step is that although truth itself is not definable, by Tarski's theorem, the property of $S$ that it satisfies the inductive definition of truth has complexity merely $\Delta_1(S)$. Thus, one of the assertions about $S$ that is true in $M[H]$ is that it satisfies the Tarskian definition of truth. Thus, there must be a condition forcing that $\dot{A}$ is a satisfaction class, obeying Tarski's inductive definition.
But since the satisfaction class of $M$ is unique, this means that there is no choice for the generic filter whether or not to place a pair into or out of $\dot{A}$. Thus, inside $M$ by simply looking at which pairs can be added at all to the class named by $\dot{A}$, we can define $S$ in $M$. But this contradicts Tarski's theorem on the non-definability of truth. QED
Let me argue next that we don't actually need the satisfaction class to be unique, and the same argument will work whenever $M$ admits a satisfaction class at all, with GBC in $(M,S)$. This can conceivably happen in non-standard models $M$, with non-standard Gödel codes. Nevertheless, the standard part of the satisfaction class, that is, for standard Gödel codes, will be unique, and so we can still get that stable part of the satisfaction class from the name $\dot{A}$---the pairs that are forced into $\dot{A}$ by every condition. This will still be a satisfaction predicate, contrary to Tarski's theory of truth, even if $M$ is not an $\omega$-model and possibly admits several satisfaction classes.
Finally, let me point out that not every model of ZFC admits a satisfaction class. For example, if $M$ is pointwise definable, as in our joint paper on Pointwise definable models of set theory, then this property would be revealed by the satisfaction class, and so we cannot add this class without revealing to $M$ that it is countable.