(As pointed out by @PaulBlainLevy, the following doesn't meet the requirement that it should have no known reduction even under large cardinal assumptions. But I think it's a natural $\Pi^2_2$ statement, so I'll leave it here.)
Consider the statement "For every set of reals $X$, $X^\#$ exists". (Equivalently, "for every set of reals $X$, there is an elementary embedding $L(\mathbb{R},X)\to L(\mathbb{R},X)$".) I claim it's $\Pi^2_2$ but not $\Sigma^2_2$, at least assuming the consistency of ZFC + "For every set of reals $X$, $X^\#$ exists". (Here I mean that there is a fixed $\Pi^2_2$ formula $\psi$ such that ZFC proves "$\psi$ holds iff $X^\#$ exists for all sets of reals $X$", but this is not the case for $\Sigma^2_2$).
It's $\Pi^2_2$: For given $X$, it is $\Sigma^2_1(\{X\})$ to say that $X^\#$ exists, as $X^\#$ is coded by a real, and one just has to check that for each countable ordinal $\alpha$, the model generated from $\mathbb{R}\cup\alpha$-many indiscernibles is wellfounded, to know that it is correct, and this is all expressed as a projective statement about some set of reals coding everything.
It's not $\Sigma^2_2$ (modulo the consistency mentioned above): For suppose it is, and fix a $\Pi^2_1$ formula $\varphi$ such that ZFC proves that $\exists A\subseteq\mathbb{R}\varphi(A)$ iff $X^\#$ exists for all sets of reals $X$. Assume ZFC + $X^\#$ exists for all sets of reals $X$. Let $A$ witness the $\Sigma^2_2$ statement, and let $A'=(A,W)$ where $W$ is a wellorder of $\mathbb{R}$. Consider $M=L(\mathbb{R},A')$. Then $M\models$ZFC, and $A\in M$, and $\mathbb{R}\subseteq M$, so note the truth of $\varphi(A)$ goes down to $M$. So $M\models\mathrm{ZFC}+V=L(\mathbb{R},A')$+$\exists A\subseteq\varphi(A)$, so models "$(A')^\#$ exists", but this is a contradiction.