(Nonconnective algebraic) K-theory cohomology = K-theory of cohomology? Situation
Suppose that we have

*

*a commutative ring (or an $E_{\infty}$-ring) $R$ and

*a homotopy type $X$.

Then we get a canonical morphism
$$
f \colon K(R ^ {\Sigma ^ \infty X_+}) \to K(R) ^ {\Sigma ^ \infty X_+}
$$
where $K$ denotes the nonconnective algebraic $K$-theory.
Question
Is $f$ an equivalence for example when $X$ is finite?
 A: The answer is typically going to be no.
Suppose for example that $R$ is discrete. Then $R^{\Sigma^\infty_+X}= C^*(X;R)$ is the usual cochain algebra, and perfect modules over that algebra are equivalent (via the homotopy fixed points functor) to the thick full stable subcategory of $Fun(X, Perf(R))$ generated by $R$ as a trivial local coefficient system.
Let's further assume that $R$ is regular noetherian so that $Perf(R)$ has the (bounded) Postnikov $t$-structure, and $Fun(X,Perf(R))$ has the pointwise $t$-structure. This pointwise $t$-structure is inherited by the full subcategory described above. Indeed, I claim that under these assumptions, a local system $L$ is in this thick subcategory if and only if each homotopy group $\pi_kL$, as a $\pi_1(X)$-module, has a filtration whose associated graded's are trivial $\pi_1(X)$-modules - this condition is obviously preserved under truncations.
It follows that the heart of this $t$-structure on this full subcategory is exactly finitely generated $R$-modules, so by the theorem of the heart, $K(C^*(X;R)) = K(R)$, where of course $K(R)^{\Sigma^\infty_+X}$ is usually not $K(R)$ (for instance if $X = S^n$, then $\pi_0$ of the former has a summand isomorphic to $\pi_n$ of the latter)
