$\DeclareMathOperator\SL{SL}\DeclareMathOperator\Sp{Sp}$Suppose that $F/Q$ is a number field.

Using automorphic forms, Borel computed the ($R$-) stable cohomology of $\SL_n(O_F)$, and as a result, computed $K_i(O_F)\otimes Q$. Nowadays, using work of Suslin, Voevodksy, Rost, etc, one has an almost "complete" understanding of the actual integral groups $K_i(Z)$, say, modulo Vandiver's conjecture. This does not *directly* give the stable cohomology of $\SL_n$, because one set of groups is computing homotopy and the other is computing cohomology, but let us not worry about that distinction for the moment.

Borel also computed the ($R$-) stable cohomology of $\Sp_{2n}(O_F)$. My question, loosely speaking, is whether one can describe the stable integral cohomology of $\Sp_{2n}(Z)$ in as detailed away as the algebraic K-groups of $Z$ "describe" the integral cohomology of $\SL_n(Z)$.

Let me summarize some of what I have found out (following up some of the answers below), mostly though emails from experts.

For affine objects, which certainly includes $Z$, K-theory is about the monoidal category $P(A)$ of projective finitely generated A-modules, and Hermitian K-theory is about the monoidal category $P(A)_h$ of objects in $P(A)$ equipped with a non-degenerate symmetric (or skew-symmetric) form. One of the issues with computing or working with such a theory over $Z$ is that irritating issues arise in characteristic $2$, as one might expect with quadratic forms present. It seems that one might be in good shape to understand the groups $K^h_i(Z[1/2])$. For usual K-theory, there is an excision formula relating $K_i(Z)$ to $K_i(Z[1/2])$ and $K_i(F_2)$. The latter group is "easy" (or at least was computed by Quillen).

Of interest to me in $K_i(Z)$ are the Soulé classes. The next thing I will be thinking about is whether Soule classes can give rise to elements in the stable cohomology of $\Sp_2n(Z)$.

Andy said some very interesting things, but I will probably be awarding the bounty to Oscar, since the paper he linked to was more directly relevant to what I was trying to find out. (Sorry Andy...but it looks like you have lots of reputation anyway!)