The additional point is the degree $n$ of the splitting field $E$ of the polynomial $g$. Since we are in characteristic $0$ and $E/\mathbf Q$ is normal, $E$ is galois of degree dividing $d$. Its Galois group $G$ permutes transitively the prime ideals $P$ of $R$ above $p$. It follows classically that all the $r$ primes $P$ of $E$ over $p$ have the same indices of ramification $e$ and of inertia $f$, so that $n=ref$. In Galois theoretic terms, let $I \subset D$ resp. be the inertia and decomposition subgroups of a prime $P$ above $p$. Recall that $s \in D$ iff $s(P)=P$, and $I$ is the kernel of the surjective homomorphism constructed as follows: given another prime $Q$ above $p$, by the definition of $D$, $x\equiv y$ mod $P$ implies $s(x)\equiv s(y)$ mod $Q$ for any $s\in D$, which allows to define naturally the quotient map $s\in D \to\bar s \in Gal((R/P)/\mathbf F_p)$ (the Galois group of the extension of residue fields). Obviously $G$ has order $n$, the $G$-orbit of $P$ has cardinal $r$, and it is classically known that $I$ has order $f$ (the proof is not absolutely immediate, see e.g. P. Samuel's ANT, chap. 66, §2), i.e. $R/P$ is a finite field of cardinal $p^f$. Note that this means that the residual polynomial $\bar g$ splits in $R/P$ because the residual extension $(R/P)/\mathbf F_p$ is galois.
You suppose moreover that $p$ does not divide the discriminant $D(K)$ (your notation $D_g$ is somewhat unusual), which is to say $p$ is unramified in $K$, hence also in $E$, so the inertia index $f$ is $1$. This means that $\bar g$ splits in $\mathbf F_p$ in this case.