Is there any first order theory whose model is not a structure within ZFC and it is within ZFC + some extra axiom A?
P.S. To define a structure within set theory ZFC+A consider the following example. Given a theory T whose only axiom is $\forall xP(x)$. Then by T has a model that is a structure in ZFC+A I mean the sentence $\exists t\exists z(R(z,t)\&\exists y\forall(x\in y)(<x>\in z))$ is a theorem of ZFC+A, where $R(z,t)$ means $z$ is an unary relation on $t$. The cases with more axioms are defined by analogy.
Sure - ZFC itself is an example. Since ZFC proves the soundness theorem, if ZFC were to prove "ZFC has a model" then ZFC would prove its own consistency, contradicting Godel's (second) incompleteness theorem. Large cardinal principles (or rather, the theories gotten by adding them to ZFC) provide further examples in increasing consistency strength.
(As an aside, ZFC also proves the completeness theorem: instead of talking about existence of models, we can just talk about consistency, which is often more concrete. That's not an issue here, but it's worth noticing.)
Preemptively addressing some common confusions:
Note that "ZFC doesn't prove "ZFC has a model"" does not mean "ZFC proves "ZFC doesn't have a model."" In particular, language like "doesn't have a model in ZFC" is dangerously ambiguous: "doesn't (have a model in ZFC)" is correct, but "(doesn't have a model) in ZFC" isn't.
By the completeness theorem, the above means that ZFC + "ZFC has no model" has a model. It turns out that the non-finite-axiomatizability of ZFC (+ the reflection theorem) shows meanwhile that for every $M\models$ ZFC, there is some $N\in M$ such that $N\models$ ZFC; the apparent contradiction is resolved by the fact that $M$'s version of ZFC will consist of ZFC + extra "nonstandard axioms" (so non-finitely-axiomatizable theories pose some genuine issues here, contra the last sentence of your answer).