Suppose$\DeclareMathOperator\SL{SL}\DeclareMathOperator\Sp{Sp}$Suppose that F/Q$F/Q$ is a number field.
Using automorphic forms, Borel computed the (R$R$-) stable cohomology of SL_
n(O_
F)$\SL_n(O_F)$, and as a result, computed K_
i(O_
F) tensor Q$K_i(O_F)\otimes Q$. Nowadays, using work of Suslin, Voevodksy, Rost, etc, one has an almost "complete" understanding of the actual integral groups K_
i(Z)$K_i(Z)$, say, modulo Vandiver's conjecture. This does not directly give the stable cohomology of SL_
n$\SL_n$, 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$R$-) stable cohomology of Sp_
2n(O_
F)$\Sp_{2n}(O_F)$.
My question, loosely speaking, is whether one can describe the stable integral cohomology of Sp_
2n(Z)$\Sp_{2n}(Z)$ in as detailed away as the algebraic K-groups of Z$Z$ "describe" the integral cohomology of SL_
n(Z)$\SL_n(Z)$.
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$Z$, K-theory is about the monoidal category
P(A)$P(A)$ of projective finitely generated A-modules, and
Hermitian K-theory is about the monoidal category
P(A)_
h$P(A)_h$ of objects in P(A)$P(A)$ equipped with a non-degenerate
symmetric (or skew-symmetric) form. One of the issues with computing or working with such a theory over Z$Z$ is that irritating issues arise in characteristic 2$2$, 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])$K^h_i(Z[1/2])$. For usual K-theory, there is an excision formula relating K_
i(Z)$K_i(Z)$ to K_
i(Z[1/2])$K_i(Z[1/2])$ and K_
i(F_2)$K_i(F_2)$. The latter group is "easy" (or at least was computed by Quillen).
Of interest to me in K_
i(Z)$K_i(Z)$ are the SouleSoulé 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)$\Sp_2n(Z)$.
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!)