Yes. This is obvious if there are no such cardinals. (I assume that the natural numbers of the universe of sets are the true natural numbers. Otherwise, the answer is no, and there is not much else to do.) 

Assume now that there are such cardinals, and that "large cardinal axiom" is something reasonable (so, provably in $\mathsf{ZFC}$, the relevant cardinals are weakly inaccessible). If $\kappa$ is the least ordinal such that $L_\kappa $ is a model of $\mathsf{ZFC}$, then $L_\kappa$ and $V$ agree on all arithmetic statements, since they share the same natural numbers, but there are no such cardinals in $L_\kappa$. Notice that there is such an ordinal $\kappa$, since any weakly inaccessible cardinal $\tau$ of $V$ (such as the large cardinal in question) is strongly inaccessible in $L$ and therefore $L_\tau\models\mathsf{ZFC}$. On the other hand, the minimality of $\kappa$ implies that no ordinal smaller than $\kappa$ is weakly (and therefore strongly) inaccessible in $L_\kappa$. ($\kappa$ itself is "tiny": it is countable in $L$.)

Now, if instead of weak inaccessibility you assume that consistency statements or the like qualify as large cardinal axioms, then "of course" the answer is no, as $\lnot A$ would be a false arithmetic statement.

[Note that H. Friedman's results, providing examples of combinatorial statements of low-level arithmetic complexity whose validity depends on large cardinals, assume more than the negation of the large cardinal in question in the "bad" case, where the nice combinatorial statement fails. The models of set theory where this happens are in fact not $\omega$-models.]