This is not a full answer, but at least it can be proved that the parity is kept using few basic properties of homology of cubical complexes.

Let $S_1$ and $S_2$ be two surfaces bounded by $C$.
The symmetric difference $S$ of $S_1$ and $S_2$ is a cycle in $\mathbb Z_2$-chain complex of the grid. [If you are not familiar with homology, think of $S$ as a multicomponent possibly self-intersecting $2$-surface with empty (!) boundary.]
The fact that $S_1$ and $S_2$ have the same parity is equivalent with showing that the number of squares forming $S$ is even.

Since $S$ is a cycle and the homology of the cubical grid is trivial, it is also a boundary of some $3$-chain over $\mathbb Z_2$. That is, there are cubes $Q_1, \dots, Q_k$ such that the boundary of the union $Q$ of these cubes is $S$. Start removing these cubes from $Q$ one by one. (That is, make the symmetric differences.) In each single removal, the parity of the number of squares of the boundary does not change, since a single cube has an even number of squares. After removing all the cubes, the boundary is empty. Therefore the boundary of $Q$ has to consist of an even number squares.