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. (See also the last paragraph "After edit" discussing how to determine this parity.)
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.
After edit: This homology approach should essentially also answer whether a given curve $C$ bounds an even or an odd area. It is just necessary to find a 2-chain $S'$ such that the boundary of $S'$ is $C$, perhaps not a minimal one. (In some bounded subcomplex of the grid containing $C$ in order to make the computations finite.) This should be then just some matrix manipulation.