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) tensor 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 Soule 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!)