The first observation is that $U$ is representable in $CAlg(R)$ by $F(S)$ (with $F$ as in your question): $$Map_{CAlg(R)}(F(S),A) \simeq Map_{SMod}(S,A) \simeq U(A).$$
By some appropriately flowery version of Yoneda's Lemma, it follows that $$ Hom(U,U) \simeq Map_{CAlg(R)}(F(S),F(S)) \simeq Map_{SMod}(S,F(S)) \simeq R \wedge \bigvee_n B\Sigma_{n+}.$$ Applying $\pi_0$ to this, is easy to see that the operation corresponding to $1 \in R_0(B\Sigma_n)$ will be precisely the classic $n$th power operation, when applied to the $R$--algebra $A = Map(\Sigma^{\infty}_+X,R)$.
Added later by request ...
Suppose given $f \in \pi_0(F(R))$. Tracing through my equivalences, the associated operation $\theta_f: \pi_0(A) \rightarrow \pi_0(A)$, for $A$ a commutative $R$--algebra, is as follows.
Firstly $f$ can regarded as an $R$-module map $f:R \rightarrow F(R)$. Similarly, $x \in \pi_0(A)$ can be regarded as an $R$-module map $x:R \rightarrow A$. Then $\theta_f(x)$ is the composite $$ R \rightarrow F(R) \rightarrow F(A) \rightarrow A,$$ where the first map is $f$, the next is $F(x)$ and the the last is the structure map for the $R$--algebra: the wedge over $n$ of the maps $A^{\wedge n}_{h \Sigma_n} \rightarrow A$.
For the $n$th power operation, recall that $\displaystyle F(R) = R \wedge \bigvee_m B\Sigma_{m+}$, and let $f_n$ be the composite $$S^0 \rightarrow B\Sigma_{n+} \rightarrow R \wedge B\Sigma_{n+}\hookrightarrow F(R).$$ Then $\theta_{f_n}$ will be the $n$th power operation. You can now specialize to $A = Map(X,R)$ if you want.