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.