No, you cannot prove this because it is not true.  For example, consider $M$ to be the product of two oriented, complete Riemannian surfaces $M=\Sigma_1\times\Sigma_2$ where $g$ is the product metric from the metric $g_i$ on $\Sigma_i$.  Let $J_i$ be the $g_i$-compatible complex structure on $\Sigma_i$ that induces the given orientation on $\Sigma_i$ and let $I = J_1 + J_2$ and $J = J_1 - J_2$.  Then $I$ and $J$ are $g$-parallel and $g$-compatible almost complex structures on $M$.  However, except in the very special situation that each $g_i$ is flat, $g$ is not hyperKähler since it is not Ricci-flat.