Is every bounded face of a closed path parameterizable? Suppose $\gamma:[0,1]\to\mathbb{R}^2$ is a closed path, and let $D$ denote some bounded component of $\gamma^c$ (writing "$\gamma$" for the trace of $\gamma$ as well).  Must there be a parameterization for $\partial D$?  
Related: Is there a characterization for what $\gamma^{-1}(\partial D)$ can look like in $[0,1]$?
EDIT:
To say that $\partial D$ is parameterizable is to say that there is a continuous surjective map from $[0,1]$ to $\partial D$.
 A: First, one needs to specify the meaning of "there exists a parameterization of $X$". In the context of your question, the only meaningful interpretation of this is that there exists a continuous surjective mapping $[0,1]\to X$. With this interpretation the answer to your question is positive. The key to the proof is
Theorem. (Hahn and Mazurkiewicz, see e.g. theorem 31.5 in Willard's book "General Topology") A compact metrizable space $X$ is a continuous image of the interval $[0,1]$ if and only if $X$ is connected and locally connected. 
In your situation, $\gamma([0,1])$ is connected and locally connected, from which one concludes that the boundary $\partial A$ of each component $A$ of ${\mathbb R}^2 - \gamma([0,1])$ satisfies the same property. Therefore, $\partial A$ is parameterizable. 
Edit. Here is an additional argument used to conclude local connectivity of $\partial A$. The argument is taken from my answer here. Set $K:=\gamma([0,1])$. It is a connected, locally connected subset of $S^2= R^2 \cup \{\infty\}$. Therefore, every component of $S^2 - K$ is simply connected. The subset $A$ is one of these components. Its complement, $A^c=S^2 - A$, is the union of $K$ with some open subsets of $S^2$. Open subsets of $S^2$ are, of course, locally connected. Since $K$ is also locally connected, so is $A^c$. Now, the Caratheodory-Torhorst extension theorem quoted in my answer implies that $\partial A$ is also locally connected. 
