= Solution
Let $\pi$ be a <prime element> of the <Gaussian integers>. Its <norm> $\pi\overline\pi$ is a positive integer greater than one, so $\pi$ divides some ordinary positive <prime number> in its integer prime factorization. It cannot divide two distinct such primes $p,q$: an integer Bézout identity $up+vq=1$ would imply $\pi\mid1$, contrary to its being a nonunit. This proves existence and uniqueness of the rational prime lying below $\pi$.
Up to associates, the complete list of Gaussian <prime elements> is: $1+i$; the positive rational primes $p\equiv3\pmod4$; and, for every rational prime $p\equiv1\pmod4$, the two nonassociate elements $a+bi$ and $a-bi$, where $a,b>0$ and $p=a^2+b^2$. Representatives related by multiplication by one of $\pm1,\pm i$ are associates. The last pair accounts for both factors of a split prime.
Every residue class modulo $p$ has a unique representative $a+bi$ with $0\leq a,b<p$, since $p\mid(a-a')+(b-b')i$ means both coordinate differences are divisible by $p$. Hence $\boxed{|R/pR|=p^2}$ and
$$
R/pR\cong\mathbb F_p[X]/(X^2+1).
$$
For $p=2$, the class of $1+i$ is nonzero but its square is $2i=0$; thus the <quotient ring> is not a <field>. For $p\equiv3\pmod4$, a root $r^2=-1$ would imply $r^{p-1}=(-1)^{(p-1)/2}=-1$, contradicting <Fermat's little theorem>. The quadratic $X^2+1$ is therefore irreducible and its quotient is a <field> with $p^2$ elements.
For $p\equiv1\pmod4$, choose a generator $g$ of the cyclic group $\mathbb F_p^\times$. Then $r=g^{(p-1)/4}$ satisfies $r^2=-1$. The two nonzero classes $i-r$ and $i+r$ have product zero in $R/pR$. Thus
$$
\boxed{R/pR\text{ is a field exactly for rational primes }p\equiv3\pmod4.}
$$
For $p\equiv1\pmod4$ it is instead isomorphic to $\mathbb F_p\times\mathbb F_p$, by the distinct linear factors and the <Chinese remainder theorem>.
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