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. Note that finiteness of the map $X \to Y$ is not necessary here.

For another argument, note that we can reduce to the case where $Y$ is $\mathrm{Spec} R$ (by making a base-change via $\mathrm{Spec} R \to Y$). 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 $\mathrm{Spec} R$. Here one uses the fact that a finite etale morphism of local rings with the same residue field must be an isomorphism, which follows from Nakayama's lemma and since flatness implies injectivity (for local rings).

(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 $S \to X$.)

(P. P. S. One doesn't need finiteness even in the second argument: if $X \to Y$ is an etale morphism with $Y$ the spectrum of a henselian ring $R$, then the local ring of a point of $x$ lying above the closed point of $Y$ is finite over $R$, by Zariski's Main Theorem: one characterization of henselian rings is that a quasi-finite local algebra over a henselian local ring is finite.)

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. Note that finiteness of the map $X \to Y$ is not necessary here. This argument seems to require modification because it is not obvious that the map will factor through the henselianization.

For another argument, note that we can reduce to the case where $Y$ is $\mathrm{Spec} R$ (by making a base-change via $\mathrm{Spec} R \to Y$). 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 $\mathrm{Spec} R$. Here one uses the fact that a finite etale morphism of local rings with the same residue field must be an isomorphism, which follows from Nakayama's lemma and since flatness implies injectivity (for local rings).

(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 $S \to X$.)

(P. P. S. One doesn't need finiteness even in the second argument: if $X \to Y$ is an etale morphism with $Y$ the spectrum of a henselian ring $R$, then the local ring of a point of $x$ lying above the closed point of $Y$ is finite over $R$, by Zariski's Main Theorem: one characterization of henselian rings is that a quasi-finite local algebra over a henselian local ring is finite.)

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 $S \to X$.)

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. Note that finiteness of the map $X \to Y$ is not necessary here.

For another argument, note that we can reduce to the case where $Y$ is $\mathrm{Spec} R$ (by making a base-change via $\mathrm{Spec} R \to Y$). 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 $\mathrm{Spec} R$. Here one uses the fact that a finite etale morphism of local rings with the same residue field must be an isomorphism, which follows from Nakayama's lemma and since flatness implies injectivity (for local rings).

(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 $S \to X$.)

(P. P. S. One doesn't need finiteness even in the second argument: if $X \to Y$ is an etale morphism with $Y$ the spectrum of a henselian ring $R$, then the local ring of a point of $x$ lying above the closed point of $Y$ is finite over $R$, by Zariski's Main Theorem: one characterization of henselian rings is that a quasi-finite local algebra over a henselian local ring is finite.)