Is the forgetful functor $\mathrm{Mod}_R \mathrm{Sp} \rightarrow \mathrm{Sp}$ faithful?

$U_R$ obviously preserves delooping, so if that were the case, because $\pi_0 map(X,Y) = \pi_1 map(X, \Sigma Y)$, you would also get an isomorphism on $\pi_0$, so an equivalence of mapping spaces.

In other words, $U_R$ is faithful if and only if it is fully faithful. But now for a map of ring spectra $R\to S$, the forgetful $Mod_S \to Mod_R$ is fully faithful if and only if $R\to S$ is an epimorphism of ring spectra (good examples are localizations - be careful that classical examples such as $R\to R/I$ for a usual ring $R$ tend to fail).

This is to say that "being an $S$-module" becomes a property of an $R$-module, rather than additional structure - so of course you can expect that to be very rare.

In your example of $H\mathbb Z$, it doesn't hold at all - you can for instance detect it on the level of the ring of stable cohomology operations of singular cohomology, which is bigger than just $\mathbb Z$ (look at the (co)homology of Eilenberg-MacLane spaces)

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.