I'll suppose $R$ is an integral domain, and therefore contained in an algebraically closed field $K$ (one can do without this assumption, but it simplifies the argument). Suppose $p(s_1,\ldots,s_n)=0$ but $p$ is not the zero polynomial; then certainly there exist constants $a_1,\ldots,a_n\in K$ such that $p(a_1,\ldots,a_n)\neq0$. Now consider $ Q=X^n+a_1X^{n-1}+a_2X^{n-2}+\cdots+a_{n-1}X+a_n\in K[X]. $ Since $K$ is algebraically closed, this polynomial splits $Q=\prod_{i=1}^n(X-r_i)$ for roots $r_1,\ldots,r_n\in K$. But that means that if $f:K[x_1,\ldots,x_n]\to K$ is the ring morphism of substituting $-r_j$ for $x_j$, for $j=1,\ldots,n$, then $a_i=f(s_i)$ for $i=1,\ldots,n$. Now $ 0=f(0)=f(p(s_1,\ldots,s_n))=p(f(s_1),\ldots,f(s_n))=p(a_1,\ldots,a_n), $ a contradiction. The before-last equality is simply the fact that a ring morphism $f$ commutes with the ring operations used in forming a polynomial expression $p$.