Some thoughts.
There are no such varieties when S = 1. This is a consequence of a theorem of Fontaine, MR1274493 (Schémas propres et lisses sur Z).
I think that one should only expect finitely many such varieties for any fixed S. Let me give an argument that uses every possible conjecture I know. There may be an unconditional proof, but that would probably require knowing something about K3-surfaces.
I first want to claim that the ramification at primes q|S is "bounded" independently of X. The corresponding fact for elliptic curves will be that the power of the conductor for each q|N is bounded by 2 (if p > 3) or (if p = 2 or 3) by some fixed number I can't remember.
The most obvious argument along these lines is to consider the representation on inertia I_q acting on the p-adic etale cohomology groups H^2(X). These correspond to Galois representations with image in GL_22(Z_p). The argument I have in mind for elliptic curves works directly in this case, providing that one has "independence of p" statement for the Weil-Deligne representations at q (quick hint: the image of wild inertia divides the gcd of the orders of GL_22(F_p) over all primes p). This may require the existence of semi-stable models, which one certainly has for elliptic curves, but I don't know for K3-surfaces.
The next step is to use a Langlands-type conjecture. The p-adic representation V on H^2(X) may be reducible, but at least we know that each irreducible chunk will correspond to an irreducible Galois representation of Q into GL_n(Z_p) for some n (at most 22). Each of these, conjecturally, will correspond to a cuspidal automorphic form of fixed weight and level divisible only by q|S. Moreover, from the previous paragraph, the level will be bounded at q|S. Thus there will only be finitely many representations which can occur as H^2(X) for any K3-surface X/Z[1/S]. (Maybe I am assuming here that the Galois representation acting on H^2(X) is semi-simple --- let us do so, since this is a conjecture of Grothendieck and Serre.)
Finally, I want to deduce from any equality H^2(X) = H^2(X') that X is (essentially) X'.
From the Tate conjecture we deduce the existence of correspondences X~~>X' and X'~~>X over Q whose composition induces an isomorphism on H^2(X) --- and now hopefully some knowledge of the geometry of K3 surfaces is enough to show that these sets of "isogenous" K3 surfaces form a finite set.
EDIT:
As Buzzard points out, I obscured the fact in the last paragraph that some more arithmetic may be necessary. What I meant to say is that understanding isogeny classes of K3's over Q will first require understanding isogeny classes over C, and hopefully this second task will be the hard part.
As David points out, the Torelli theorem for K3 will surely be relevant here. I think there can be non-isomorphic isogenous K3s, however. If one takes an isogeny of abelian surfaces A->B then one can presumably promote this to an isogeny of the associated Kummer surfaces.
EDIT:
Here is another thought. Deligne proves the Weil conjecture for K3 surfaces:
http://www.its.caltech.edu/~clyons/DeligneWeilK3trans.pdf
The philosophy is that there should be an inclusion of motives H^2(X) --> H^1(A) tensor H^1(A) for some abelian variety A (possibly of some huge but uniformly bounded dimension, like 2^19). It may be possible (conjecturally or otherwise) to reduce your question to the analogous statement for A, for which it is known. (Prop 6.5 is relevant here). It may well be possible to show that the variety A is defined over Z[1/2S], for example. I could make this edit more coherent but I'm off to lunch, so treat this as a thought fragment.
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