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Here is an argument. Use with caution; it's possible that I've made a mistake. We don't use any properties of $\Sigma^\infty BK$ -- this could be an arbitrary spectrum. But it's crucial that that we use $\Sigma^\infty BS$ with $S$ a finite $p$-group.

1. Note that if $S$ is a finite $p$-group, then its rational group cohomology (with trivial coefficients) vanishes. To see this, we reduce to the case where $S$ is abelian, where this is a standard calculation. For if $S$ is nonabelian, then there is always a nontrivial short exact sequence $Z(S) \to S \to S / Z(S)$ where $Z(S)$ is the center of $S$. Because $Z(S)$ is the center of $S$, the action of $S/Z(S)$ on $Z(S)$ is trivial. Thus in the Serre spectral sequence $H^\ast(B(S/Z(S)); H^\ast(B(Z(S))) \Rightarrow H^\ast(S)$ we have trivial coefficients. So we can induct on the order of $S$.

2. Let $\ell \neq p$ be a prime different from $p$. We claim that $\Sigma^\infty BS \wedge M(\ell) = 0$, where $M(\ell)$ is the mod-$\ell$ Moore spectrum. First we show this when $S = C_{p^k}$. For this, it suffices to observe that $\ell: H_\ast(B C_{p^k};\mathbb Z) \to H_\ast(B C_{p^k};\mathbb Z)$ is an isomorphism, so that $\Sigma B C_{p^k} \wedge M(\ell)$ is contractible. Then we can induct on the order of $S$ using the same exact sequence $Z(S) \to S \to S/Z(S)$ from before, applying the $M(\ell)$-based Atiyah-Hirzebruch spectral sequence $H_\ast(B(S/Z(S));\underline{H_\ast(B(Z(S));M(\ell)_\ast)}) \Rightarrow M(\ell)_\ast(BS)$ where again the $\pi_1$-action is trivial.

What (2) really says is that $\Sigma^\infty BS$ is $p$-local.

3. Let $X$ be such that $X \wedge M(p) = 0$. We claim that $Map(X,F(\Sigma^\infty BS, \Sigma^\infty BK)) = 0$. Equivalently, the claim is that $Map(\Sigma^\infty BS, F(X,\Sigma^\infty BK)) = 0$. By (2), it is equivalent to claim that $Map(\Sigma^\infty BS, F(X,\Sigma^\infty BK)_{(p)}) = 0$, where we have taken a $p$-localization. But since $X \wedge M(p) = 0$, we actually have that $F(X,\Sigma^\infty BK)_{(p)}$ is rational. Now, every rational spectrum splits as a sum of shifts of $H\mathbb Q$, so it suffices to show that $F(\Sigma^\infty BS, H\mathbb Q) = 0$, i.e. that $H^\ast(BS;\mathbb Q) = 0$, which is precisely (1).