The Adjacency Matrix of Symmetric Differences of any Subset of Faces has an Eigenvalue of $2$...? Assume a cubic (for the bounty) planar graph $G$ and let's call its faces $f_k\in F$. The adjacency matrix of any face $f_k$ has an eigenvalue of $2$, since it's a $2$-regular graph, i.e. a cycle.
I want to show that the symmetric differences of any subset of $F$, also has an eigenvalue of $2$, which is obvious since $f_k \ominus f_n$ is again a cycle.
So I started to setup a symbolic calculation:
$$
A(\vec x)={\huge{\ominus}}_{k} \; x_k f_k = \left( \sum_k x_k f_k \right) \bmod 2  ,
$$
where $\vec x$ denotes a vector with values from $\mathbb Z_2$, i.e. $0$ and $1$. Once $\vec x$ is chosen the resulting matrix contains only values $0$ and $1$.
Now I want to show that $\displaystyle \det\left(A(\vec x) -2I \right)=0 \; \; \forall \vec x$, except $\vec 0$ and $\vec 1$ (where all $x_k=1$). And this is were I'm stuck...
To give an explicite example, here the resulting matrix for the triangular prism graph $Y_3$
$\hskip1.7in$
$$
A(\vec x)=
\pmatrix{       0& x_1+x_3& x_1+x_5& x_3+x_5&       0&       0\\
 x_1+x_3&       0& x_1+x_4&       0& x_3+x_4&       0\\
 x_1+x_5& x_1+x_4&       0&       0&       0& x_4+x_5\\
 x_3+x_5&       0&       0&       0& x_2+x_3& x_2+x_5\\
       0& x_3+x_4&       0& x_2+x_3&       0& x_2+x_4\\
       0&       0& x_4+x_5& x_2+x_5& x_2+x_4&       0
}
$$
where the triangles are easily recognized as upper left/lower right blocks with variables $x_1/x_2$.
Now when calculating $\det\left(A(\vec x) -2I \right)=0$ for all $32-2=30$ possibilities, I can verify that this is correct!
 A: The best way to prove for cubic graphs is to notice that any symmetric difference of faces is $2$-regular.  The argument here is basically that any two faces either share $0$ or $1$ common edge, and so the symmetric difference is either two disjoint cycles or a bigger cycle.  Then you know that a two-regular graph has an eigenvalue of $2$.
You can do it the way you are trying to; it is difficult.  The fact that you are considering adding matrices $\bmod{2}$, then converting back to matrices over $\mathbb{Q}$ makes things hard.  But it can help if you use the $\circ$ operation, basically $A \circ B$ is the componentwise multiplication.  For $0$-$1$ matrices this means you have a $1$ in positions where $A$ and $B$ are both $1$, and a $0$ everywhere else.  This means the symmetric difference of $A$ and $B$, $A+B \bmod{2}$, can be written $A+B - 2(A \circ B)$.  So you can try to show for two faces: take $F_{i}$ and $F_{j}$, considered as the adjacency matrices of two faces (as $n \times n$ matrices, $|V|=n$).  You know each face has an eigenvector for $2$, $f_{i}$ and $f_{k}$, respectively where these vectors have a $1$ corresponding to points on the face and a $0$ everywhere.  Then you need to convince yourself that
$$(F_{i}+F_{j} - 2(F_{i} \circ F_{j})) (f_{i} + f_{j} - 2(f_{i} \circ f_{j}))=  2(f_{i} + f_{j} - 2(f_{i} \circ f_{j})).$$
This is really easy if $F_{i}$ and $F_{j}$ don't share any common edges (because then $F_{i} \circ F_{j} = 0$, $f_{i} \circ f_{j} = 0$, and $F_{i}f_{j} = F_{j}f_{i} = 0$).  Some more (complicated but not undoable) arguments need to be made if the two faces share an edge...
[OLD ANSWER for non-cubic graphs]
There are really two things to say: if the symmetric difference of a collection of faces is again a cycle, or even a union of disjoint cycles, then there is nothing to show; it is two regular, so clearly has an eigenvalue equal to $2$.
But is this true? What happens if two faces share a vertex but not an edge? For example, take this graph; it is planar.  The symmetric difference of the top and bottom faces is the whole graph.  But it does not have an eigenvalue of two.

