Can a curve fill a disk?

Well, note that $\gamma$ being $C^1$ implies that $\gamma$ is rectifiable. Hence, the Hausdorff dimension of the image of $\gamma$ is $1$ (unless, of course, $\gamma$ is constant). However, the Hausdorff dimension of the unit disk is $2$, by the existence of the Lesbegue measure.

The image of a $C^1$ map $\Bbb R \to \Bbb R^2$ has measure zero. This is Sard's theorem.