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Yes. Letting $k$ be the field of fractions of $R$, we have the following commutative diagram. $$\begin{array}{ccc} R&\stackrel{f}{\rightarrow}&S\\ \downarrow\scriptstyle{}&&\downarrow\scriptstyle{}\\ k&\stackrel{g}{\rightarrow}&S_{\mathfrak{p}} \end{array}$$ However, $f$ and the localization $S\to S_{\mathfrak{p}}$ are epimorphisms, so $g$ is an epimorphism with domain a field. This means that it is surjective, so $S_{\mathfrak{p}}$ is a field, and $\mathfrak{p}=0$.
To see that an epimorphism $f\colon g\colon k\to A$ of commutative rings with domain a field $k$ is surjective, consider the morphisms $u,v\colon A\to A\otimes_k A$ given by $u(a)=a\otimes1$ and $v(a)=1\otimes a$. Then, $u\circ g=v\circ g$ and, from the definition of epimorphism, $u=v$, in which case $g(k)=A$.
Yes. Letting $k$ be the field of fractions of $R$, we have the following commutative diagram. $$\begin{array}{ccc} R&\stackrel{f}{\rightarrow}&S\\ \downarrow\scriptstyle{}&&\downarrow\scriptstyle{}\\ k&\stackrel{g}{\rightarrow}&S_{\mathfrak{p}} \end{array}$$ However, $f$ and the localization $S\to S_{\mathfrak{p}}$ are epimorphisms, so $g$ is an epimorphism with domain a field. This means that it is surjective, so $S_{\mathfrak{p}}$ is a field, and $\mathfrak{p}=0$.
To see that an epimorphism $f\colon k\to A$ of commutative rings with domain a field $k$ is surjective, consider the morphisms $u,v\colon A\to A\otimes_k A$ given by $u(a)=a\otimes1$ and $v(a)=1\otimes a$. Then, $u\circ g=v\circ g$ and, from the definition of epimorphism, $u=v$, in which case $g(k)=A$.