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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.

Proposition: If $S$ is a finite $p$-group and $U$ is any spectrum, then $F(\Sigma^\infty BS, U)$ is $p$-complete.

We will prove this using the following lemmas:

Lemma 1: If $S$ is a finite $p$-group, then $\tilde H_\ast(BS; A) = 0$ if $p: A \to A$ is an isomorphism.

As Piotr points out in his answer, this is a consequence of Maschke's theorem. We provide a proof using basic algebraic topology and finite group theory.

Proof: 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 $H_\ast(B(S/Z(S)); \underline{H_\ast(B(Z(S))}) \Rightarrow H_\ast(BS)$ we have trivial coefficients. So we can induct on the order of $S$.

Corollary 2: Let $S$ be a finite $p$-group. Then $\Sigma^\infty BS$ is $p$-local.

Proof: Let $\ell \neq p$ be a different prime. The claim is that $\ell: \Sigma^\infty BS \to \Sigma^\infty BS$ is an equivalence of spectra. It suffices to show that $\ell: \Sigma BS \to \Sigma BS$ is an equivalence of spaces. By the homology Whitehead theorem for field coefficients, it suffices to show that $\ell: \tilde H_\ast(BS;k) \to \tilde H_\ast(BS;k)$ is an isomorphism for $k = \mathbb Q$ or $k = \mathbb F_q$ with $q$ a prime. But if $k = \mathbb Q$ or $k = \mathbb F_q$ with $q \neq p$, both sides are zero by Lemma 1, while if $k = \mathbb F_p$, then this follows from $\ell$ being prime to $p$.

Lemma 3: Let $T$ be a $p$-torsion spectrum -- i.e. $T$ is $p$-local and has trivial rationalization -- and let $U$ be an arbitrary spectrum. Then $F(T,U)$ is $p$-complete.

Proof: Let $X$ be such that $X \wedge M(p) = 0$; the claim is that $Map(X,F(T,U)) = 0$. Equivalently, the claim is that $Map(T, F(X,U)) = 0$. Because $T$ is $p$-local, it is equivalent to claim that $Map(T, F(X,U)^{(p)}) = 0$, where we have taken a $p$-colocalization (i.e. we have applied the right adjoint $(-)^{(p)}$ to the inclusion of the $p$-local spectra into all spectra). Then, $$F(X,U)^{(p)} \wedge M(p) = (F(X,U)\wedge M(p))^{(p)} = F(X \wedge \Sigma^{-1} M(p), U)^{(p)} = 0$$ The first equivalence comes because $(-)^{(p)}$ commutes with finite colimits, the second by Spanier-Whitehead duality, and the third by the hypothesis that $X \wedge M(p) = 0$. Since $F(X,U)^{(p)}$ is by definition $p$-local, this means that it is rational. So $$Map(T, F(X,U)^{(p)}) = Map_{H\mathbb Q}(H\mathbb Q \wedge T, F(X,U)^{(p)}) = 0$$ because by hypothesis $H\mathbb Q \wedge T = 0$.

Proof of Proposition: By Lemma 1, $\Sigma^\infty BS$ has trivial rationalization, and by Corollary 2, $\Sigma^\infty BS$ is $p$-local. So by Lemma 3, $F(\Sigma^\infty BS, U)$ is $p$-complete.

Here is an argument (not as clean as Piotr's below). 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.

Proposition: If $S$ is a finite $p$-group and $U$ is any spectrum, then $F(\Sigma^\infty BS, U)$ is $p$-complete.

We will prove this using the following lemmas:

Lemma 1: If $S$ is a finite $p$-group, then $\tilde H_\ast(BS; A) = 0$ if $p: A \to A$ is an isomorphism.

As Piotr points out in his answer, this is a consequence of Maschke's theorem. We provide a proof using basic algebraic topology and finite group theory.

Proof: 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 $H_\ast(B(S/Z(S)); \underline{H_\ast(B(Z(S))}) \Rightarrow H_\ast(BS)$ we have trivial coefficients. So we can induct on the order of $S$.

Corollary 2: Let $S$ be a finite $p$-group. Then $\Sigma^\infty BS$ is $p$-local.

Proof: Let $\ell \neq p$ be a different prime. The claim is that $\ell: \Sigma^\infty BS \to \Sigma^\infty BS$ is an equivalence of spectra. It suffices to show that $\ell: \Sigma BS \to \Sigma BS$ is an equivalence of spaces. By the homology Whitehead theorem for field coefficients, it suffices to show that $\ell: \tilde H_\ast(BS;k) \to \tilde H_\ast(BS;k)$ is an isomorphism for $k = \mathbb Q$ or $k = \mathbb F_q$ with $q$ a prime. But if $k = \mathbb Q$ or $k = \mathbb F_q$ with $q \neq p$, both sides are zero by Lemma 1, while if $k = \mathbb F_p$, then this follows from $\ell$ being prime to $p$.

Lemma 3: Let $T$ be a $p$-torsion spectrum -- i.e. $T$ is $p$-local and has trivial rationalization -- and let $U$ be an arbitrary spectrum. Then $F(T,U)$ is $p$-complete.

Proof: Let $X$ be such that $X \wedge M(p) = 0$; the claim is that $Map(X,F(T,U)) = 0$. Equivalently, the claim is that $Map(T, F(X,U)) = 0$. Because $T$ is $p$-local, it is equivalent to claim that $Map(T, F(X,U)^{(p)}) = 0$, where we have taken a $p$-colocalization (i.e. we have applied the right adjoint $(-)^{(p)}$ to the inclusion of the $p$-local spectra into all spectra). Then, $$F(X,U)^{(p)} \wedge M(p) = (F(X,U)\wedge M(p))^{(p)} = F(X \wedge \Sigma^{-1} M(p), U)^{(p)} = 0$$ The first equivalence comes because $(-)^{(p)}$ commutes with finite colimits, the second by Spanier-Whitehead duality, and the third by the hypothesis that $X \wedge M(p) = 0$. Since $F(X,U)^{(p)}$ is by definition $p$-local, this means that it is rational. So $$Map(T, F(X,U)^{(p)}) = Map_{H\mathbb Q}(H\mathbb Q \wedge T, F(X,U)^{(p)}) = 0$$ because by hypothesis $H\mathbb Q \wedge T = 0$.

Proof of Proposition: By Lemma 1, $\Sigma^\infty BS$ has trivial rationalization, and by Corollary 2, $\Sigma^\infty BS$ is $p$-local. So by Lemma 3, $F(\Sigma^\infty BS, U)$ is $p$-complete.

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. Now we are ready to prove the claim that $F(\Sigma^\infty BS, \Sigma^\infty BK)$ is $p$-complete.

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 $p$-colocalization (i.e. we have applied the right adjoint $(-)^{(p)}$ to the inclusion of the $p$-local spectra into all spectra). But since $X \wedge M(p) = 0$, we actually have that $F(X,\Sigma^\infty BK)^{(p)}$ is rational. So $Map(\Sigma^\infty BS, F(X,\Sigma^\infty BK)^{(p)}) = Map_{H\mathbb Q}(H\mathbb Q \wedge \Sigma^\infty BS, F(X,\Sigma^\infty BK)^{(p)}) = 0$ because by (1), $H\mathbb Q \wedge \Sigma^\infty BS = 0$.

Proposition: If $S$ is a finite $p$-group and $U$ is any spectrum, then $F(\Sigma^\infty BS, U)$ is $p$-complete.

We will prove this using the following lemmas:

Lemma 1: If $S$ is a finite $p$-group, then $\tilde H_\ast(BS; A) = 0$ if $p: A \to A$ is an isomorphism.

As Piotr points out in his answer, this is a consequence of Maschke's theorem. We provide a proof using basic algebraic topology and finite group theory.

Proof: 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 $H_\ast(B(S/Z(S)); \underline{H_\ast(B(Z(S))}) \Rightarrow H_\ast(BS)$ we have trivial coefficients. So we can induct on the order of $S$.

Corollary 2: Let $S$ be a finite $p$-group. Then $\Sigma^\infty BS$ is $p$-local.

Proof: Let $\ell \neq p$ be a different prime. The claim is that $\ell: \Sigma^\infty BS \to \Sigma^\infty BS$ is an equivalence of spectra. It suffices to show that $\ell: \Sigma BS \to \Sigma BS$ is an equivalence of spaces. By the homology Whitehead theorem for field coefficients, it suffices to show that $\ell: \tilde H_\ast(BS;k) \to \tilde H_\ast(BS;k)$ is an isomorphism for $k = \mathbb Q$ or $k = \mathbb F_q$ with $q$ a prime. But if $k = \mathbb Q$ or $k = \mathbb F_q$ with $q \neq p$, both sides are zero by Lemma 1, while if $k = \mathbb F_p$, then this follows from $\ell$ being prime to $p$.

Lemma 3: Let $T$ be a $p$-torsion spectrum -- i.e. $T$ is $p$-local and has trivial rationalization -- and let $U$ be an arbitrary spectrum. Then $F(T,U)$ is $p$-complete.

Proof: Let $X$ be such that $X \wedge M(p) = 0$; the claim is that $Map(X,F(T,U)) = 0$. Equivalently, the claim is that $Map(T, F(X,U)) = 0$. Because $T$ is $p$-local, it is equivalent to claim that $Map(T, F(X,U)^{(p)}) = 0$, where we have taken a $p$-colocalization (i.e. we have applied the right adjoint $(-)^{(p)}$ to the inclusion of the $p$-local spectra into all spectra). Then, $$F(X,U)^{(p)} \wedge M(p) = (F(X,U)\wedge M(p))^{(p)} = F(X \wedge \Sigma^{-1} M(p), U)^{(p)} = 0$$ The first equivalence comes because $(-)^{(p)}$ commutes with finite colimits, the second by Spanier-Whitehead duality, and the third by the hypothesis that $X \wedge M(p) = 0$. Since $F(X,U)^{(p)}$ is by definition $p$-local, this means that it is rational. So $$Map(T, F(X,U)^{(p)}) = Map_{H\mathbb Q}(H\mathbb Q \wedge T, F(X,U)^{(p)}) = 0$$ because by hypothesis $H\mathbb Q \wedge T = 0$.

Proof of Proposition: By Lemma 1, $\Sigma^\infty BS$ has trivial rationalization, and by Corollary 2, $\Sigma^\infty BS$ is $p$-local. So by Lemma 3, $F(\Sigma^\infty BS, U)$ is $p$-complete.

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. Now we are ready to prove the claim that $F(\Sigma^\infty BS, \Sigma^\infty BK)$ is $p$-complete.

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).

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. Now we are ready to prove the claim that $F(\Sigma^\infty BS, \Sigma^\infty BK)$ is $p$-complete.

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 $p$-colocalization (i.e. we have applied the right adjoint $(-)^{(p)}$ to the inclusion of the $p$-local spectra into all spectra). But since $X \wedge M(p) = 0$, we actually have that $F(X,\Sigma^\infty BK)^{(p)}$ is rational. So $Map(\Sigma^\infty BS, F(X,\Sigma^\infty BK)^{(p)}) = Map_{H\mathbb Q}(H\mathbb Q \wedge \Sigma^\infty BS, F(X,\Sigma^\infty BK)^{(p)}) = 0$ because by (1), $H\mathbb Q \wedge \Sigma^\infty BS = 0$.