I believe I miss something simple as I'm not quite good in foundations. Please tell me if I use some notion incorrectly.
As you know $\mathbb R$ is defined uniquely up to isomorphism. The theory of $\mathbb R$ is given by the second-order arithmetic, which is categorical and denoted by $Z_2$. The reals also can be defined in set theory. But taking into account the axiom of determinacy (AD) in fact we have two incompatible set theories: ZFC and ZF+AD. The question of existence of non-measurable sets is answered differently in these theories. And so the theories of reals in two set theories are not equivalent. Does it mean $Z_2$ is incompatible with one of them? If yes, then with which of them, and does it compatible with the other? Does it mean the unique (standard) model of $Z_2$ is also model in only one set theory?
The categoricity of $\mathbb R$ is always in reference to a fixed model of set theory. You can have the same set theory but different models thereof. In this sense I disagree with the conclusion of the exchange of comments following the question and agree with Andrés's adjective "essentially" which however is rather significant when spelled out. For example, there are models of ZF+ACC that feature nonmeasurable sets and other models that don't.
Remark by Andrés: "The issue manifests itself much earlier. For instance, (conceivably) there are models of ZF where Con(ZF) holds and models where it fails, and this is a Π$_1^0$ statement, much simpler complexity-wise than a question about measurable sets, which actually is not even within the scope of $Z_2$ (unless the nonmeasurability manifests itself fairly early)."