I think this is true. Namely, we can lift each $\mathrm{Spec}(R/\mathfrak{m}^n)\to Y$ to $X$ (because formal etaleness allows the lift to exist for any nilpotent thickening) successively so as to be compatible with all the previous ones (in fact, it has to be, by uniqueness in the lifting property). This successive lifting gives a map $\mathrm{Spec} R \to X$ lifting $\mathrm{Spec} R \to Y$ (to see this, note that without loss of generality, $X, Y$ are affine, and then taking direct limits in the category of affine schemes is the same as taking inverse limits in the category of rings). Thus we get a map from the completion $\\mathrm{Spec} \hat{R}$, hence from $\mathrm{Spec} R$. I believe this is true even if we just assume $X \to Y$ to be smooth, since then we can still make successive liftings.
For another argument, note that we can reduce to the case where $Y$ is $\mathrm{Spec} R$ (by making a base-change to $X$). Then we have to show that there is a section $\mathrm{Spec} R \to X$: this is clear because $X$ is a product of strictly henselian rings finite and etale over $R$, and if one of them also has residue field $k$, it must be isomorphic to $\mathrm{Spec} R$.
(P.S. Just in case this was the question, recall the nilpotent lifting property for an etale morphism $X \to Y$: given any scheme $S$ and subscheme $S_0$ cut out by a nilpotent ideal (one can reduce to the square-zero case by induction), any diagram with $S_0 \to X$ and $S \to Y$ leads to a unique lift $S \to X$.)