I know this is an old and already answered question, but I thought I'd elaborate a bit on what's going on "behind the scenes." When dealing with rank-two tensors, it's sometimes better not to think of matrices at all, because that blurs the distinction between something like $A_{\mu \nu}$ and something like $A^\mu_{\ \ \nu}$, which are very qualitatively different beasts. $A_{\mu \nu}$ is a bilinear form - a bilinear map that inputs two vectors and outputs a scalar, or equivalently, a linear map that inputs a vector and outputs a one-form. $A^\mu_{\ \ \nu}$ is a linear operator - a bilinear map that inputs a vector and a one-form and outputs a scalar, or equivalently, a linear map that inputs a vector and outputs another vector. They're very different, because for a bilinear form, the input and output live in two completely different spaces and can't be added or directly compared. (Think of it as being like a non-square matrix.)
If you think back to your linear algebra course, you'll probably remember that the matrices were always considered as linear operators. And in fact, many linear algebra tools, like eigenvectors, eigenvalues, trace, determinant, etc., are only defined for linear operators.
So in fact, you can't take the trace of a bilinear form $A_{\mu\nu}$. Nor can you define its eigenvalues or eigenvectors. Sure, you can write it out as a matrix in some particular basis, pretend it was a linear operator, and then find the eigenvalues of that linear operator, but the result will be basis-dependent and therefore of no physical interest. When you say you can "use the metric to take the trace $\eta^{\mu \nu} A_{\mu \nu}$ of the bilinear form $A_{\mu \nu}$," what you're really doing is (a) using the metric to convert the bilinear form into a linear operator, and then (b) taking the trace of that linear operator without using the metric. In fact, the trace of a linear operator is a metric-independent concept, because the indices are already in the right places and ready to be contracted.
(The determinant is actually a bit of a subtle case, because even though the metric $g_{\mu \nu}$ is a bilinear form, for non-Cartesian coordinate systems, the determinant $\det g$ of the metric is a valid and useful concept. It's not a tensor though, it's a "tensor density." That's another story.)
Anyway, the metric's "job" is to raise and lower indices, so if it's not actually changing index heights then it should act trivially. So the one-index-up, one-index-down form of the metric is always the identity linear operator $\delta^\mu_{\ \ \nu}$. That's true in any coordinate system, curved spacetime, whatever. That's why people almost always use the notation $\delta^\mu_{\ \ \nu}$ instead of $\eta^\mu_{\ \ \nu}$. That's also why the two-upper-index tensor $\eta^{\mu \nu}$ is defined to the inverse of the metric $\eta_{\mu \nu}$. (Again, note the subtlety - the metric inputs a vector and outputs a one-form, so the inverse metric must input a one-form and output a vector.)
So now the answer to your question is clear: the trace of the metric is always just $\delta^\mu_{\ \ \mu} = d$, the number of spacetime dimensions. Again, true in any coordinate system, any metric signature, curved spacetime, what have you. That fact that the trace of the matrix representation of $\eta_{\mu \nu}$ is 2 has no physical significance. (Well, okay, fine, in Cartesian coordinates on flat spacetime it's the summed metric signature, which is physically significant. But in general it has no physical significance.)
