It seems that the answer is "yes".

Let us identify $\mathbb{R}^2$ with $\mathbb{S}^2\setminus\{n\}$. Each Jordan curve $\gamma$ bounds a disc containing $n$. This disc admits a conformal parametrization by the unit disc $\mathbb{D}$ such that its center goes to $n$. This parametrization is unique up to rotation of $\mathbb{D}$. In particular, the image $\gamma_r$ of the circle of radius $r$ in $\mathbb{D}$ is completely determined by $\gamma$. Note that there is a homotopy from that sends $\gamma$ to $\gamma_{1/2}$.

So the question is reduced to the smooth case, and you know it already.

It seems that the answer is "yes".

Let us identify $\mathbb{R}^2$ with $\mathbb{S}^2\setminus\{n\}$. Each Jordan curve $\gamma$ bounds a disc containing $n$. This disc admits a conformal parametrization by the unit disc $\mathbb{D}$ such that its center goes to $n$. This parametrization is unique up to rotation of $\mathbb{D}$. In particular, the image $\gamma_r$ of the circle of radius $r$ in $\mathbb{D}$ is completely determined by $\gamma$. Note that there is a homotopy from that sends $\gamma$ to $\gamma_{1/2}$. The continuity at $t=1$ follows from Thm 15 (VIII, §81) in "Automorphic Functions" by Lester R Ford.

So the question is reduced to the smooth case, and you know it already.