No. $H$ is isomorphic to $G$ if and only if the same is true of their complements. The complements of $G$ and $H$ are required to be mildly sparse graphs (certainly $k$-regular for any $k$ is fine, as long as $n$ is reasonably large), and one can certainly find $k$ regular isospectral non-isomorphic graphs.
Igor Rivin
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