Say, on a complete separable metric space. Separable probably doesn't matter.
It's easy to see the opposite; if $D$ is a dense $G_\delta$ set, it's a countable intersection of open sets which contain $D$, so must also be dense. Consequently the complements of those sets are closed sets with no interior, so must be nowhere dense. And so their union (the complement of $D$) must be meager, so $D$ is comeager. Great.
It's not hard to extend this logic a bit more; if $C$ is any comeager set, then there is a $C'\subset C$ where $C'$ is comeager and $G_\delta$. But must $C'$ be dense? Must $C$ be dense?
I suppose the real question then is this: if $C\subset X$ is comeager, must $C$ be dense? (the converse definitely does not hold, see $\mathbb Q\subset\mathbb R$). By the reduction in the previous paragraph, it's enough to consider $C$ to be $G_\delta$, but that doesn't obviously lead to density. Does it?