$U_R$ is not faithful in general. Let $R=H(\mathbf{Z}/2)$. Then $\pi_*(map(R,R))$ is the Steenrod algebra $\mathcal{A}^*$ where $map$ denotes the mapping spectrum. The mapping spectrum $map(R,R)$ therefore has non-zero homotopy groups in negative degrees and differs from the mapping spectrum of $R$-module maps from $R$ to itself, which is just $R$ again, whose homotopy groups consist of $\mathbf{Z}/2$ concentrated in degree zero.

To see this difference directly in terms of mapping spaces as opposed to mapping spectra, we consider maps from $R$ to deloopings of $R$. For example, $$\pi_1(Map_{R-Mod}(R, R[2])) \cong \pi_0(Map_{R-Mod}(R, \Omega R[2])) \cong \pi_0(Map_{R-Mod}(R, R[1])) \cong \mathrm{Ext}^1_R(R,R) = 0$$ but $$\pi_1(Map_{Sp}(R,R[2])) \cong \pi_0(Map_{Sp}(R, \Omega R[2])) \cong \pi_0(Map_{Sp}(R,R[1])) = \mathcal{A}^1 \cong \mathbf{Z}/2$$ so the comparison map is not surjective on $\pi_1$.

In general, the functor $U_R$ does not induce isomorphisms on higher homotopy groups of mapping spaces. Let $R=H(\mathbf{Z}/2)$. Then $\pi_*(map(R,R))$ is the Steenrod algebra $\mathcal{A}^*$ where $map$ denotes the mapping spectrum. The mapping spectrum $map(R,R)$ therefore has non-zero homotopy groups in negative degrees and differs from the mapping spectrum of $R$-module maps from $R$ to itself, which is just $R$ again, whose homotopy groups consist of $\mathbf{Z}/2$ concentrated in degree zero.

To see this difference directly in terms of mapping spaces as opposed to mapping spectra, we consider maps from $R$ to deloopings of $R$. For example, $$\pi_1(Map_{R-Mod}(R, R[2])) \cong \pi_0(Map_{R-Mod}(R, \Omega R[2])) \cong \pi_0(Map_{R-Mod}(R, R[1])) \cong \mathrm{Ext}^1_R(R,R) = 0$$ but $$\pi_1(Map_{Sp}(R,R[2])) \cong \pi_0(Map_{Sp}(R, \Omega R[2])) \cong \pi_0(Map_{Sp}(R,R[1])) = \mathcal{A}^1 \cong \mathbf{Z}/2$$ so the induced map on $\pi_1$ is not surjective.

Let $R=H(\mathbf{Z}/2)$. Then $\pi_*(map(R,R))$ is the Steenrod algebra $\mathcal{A}^*$ where $map$ denotes the mapping spectrum. The mapping spectrum $map(R,R)$ therefore has non-zero homotopy groups in negative degrees and differs from the mapping spectrum of $R$-module maps from $R$ to itself, which is just $R$ again, whose homotopy groups consist of $\mathbf{Z}/2$ concentrated in degree zero.

To see this difference directly in terms of mapping spaces as opposed to mapping spectra, we consider maps from $R$ to deloopings of $R$. For example, $$\pi_1(Map_{R-Mod}(R, R[2])) \cong \pi_0(Map_{R-Mod}(R, \Omega R[2])) \cong \pi_0(Map_{R-Mod}(R, R[1])) \cong \mathrm{Ext}^1_R(R,R) = 0$$ but $$\pi_1(Map_{Sp}(R,R[2])) \cong \pi_0(Map_{Sp}(R, \Omega R[2])) \cong \pi_0(Map_{Sp}(R,R[1])) = \mathcal{A}^1 \cong \mathbf{Z}/2$$ so the comparison map is not surjective on $\pi_1$.

$U_R$ is not faithful in general. Let $R=H(\mathbf{Z}/2)$. Then $\pi_*(map(R,R))$ is the Steenrod algebra $\mathcal{A}^*$ where $map$ denotes the mapping spectrum. The mapping spectrum $map(R,R)$ therefore has non-zero homotopy groups in negative degrees and differs from the mapping spectrum of $R$-module maps from $R$ to itself, which is just $R$ again, whose homotopy groups consist of $\mathbf{Z}/2$ concentrated in degree zero.

To see this difference directly in terms of mapping spaces as opposed to mapping spectra, we consider maps from $R$ to deloopings of $R$. For example, $$\pi_1(Map_{R-Mod}(R, R[2])) \cong \pi_0(Map_{R-Mod}(R, \Omega R[2])) \cong \pi_0(Map_{R-Mod}(R, R[1])) \cong \mathrm{Ext}^1_R(R,R) = 0$$ but $$\pi_1(Map_{Sp}(R,R[2])) \cong \pi_0(Map_{Sp}(R, \Omega R[2])) \cong \pi_0(Map_{Sp}(R,R[1])) = \mathcal{A}^1 \cong \mathbf{Z}/2$$ so the comparison map is not surjective on $\pi_1$.