Theorem: Schinzel's hypothesis H implies the conjecture.
Proof: Choose distinct primes $q_S > 100|G|$ indexed by the 2-element subsets $S$ of $G$. For each $i \in G$, let $Q_i$ be the set of $q_S$ for $S$ such that $i \in S$ and the edge $S$ is not part of $G$. Let $P_i$ be the product of the primes in $Q_i$. Let $P = 4 \prod_S q_S^2$.
By the Chinese remainder theorem, for each $i$ we can find a positive integer $a_i$ such that
$a_i \equiv 1 \bmod{\ell^2}$ for each prime $\ell \le 10|G|$,
$a_i \equiv q-1 \bmod{q^2}$ for each $q \in Q_i$, and
$a_i \equiv 1 \bmod{q_S}$ for each $q_S \notin Q_i$.
Moreover, we can choose the $a_i$ to be distinct. Let $J$ be the set of positive integers up to $\operatorname{max} a_i$, but excluding all of the $a$'s themselves (i.e., $J$ consists of the numbers in the gaps). For each $j \in J$ choose a prime $s_j$ much larger than all the $a_i$ and all the $q_S$.
Consider the linear polynomials $P n + a_i$ and $(P n + a_i + 1)/(2P_i)$ In $\mathbf{Z}[n]$. For each prime $\ell \le 10|G|$ and each $\ell$ of the form $q_S$, all these $2|G|$ polynomials are nonzero mod $\ell$ at $n=0$. For each other prime $\ell$, there exists $n$ such that all these polynomials are nonzero mod $\ell$, since $n$ needs to avoid no more than $2|G|$ residue classes mod $\ell$. Furthermore, we can impose the condition that $P n+j$ is divisible by $s_j^2$ for each $j \in J$, and still find $n$ as above. Therefore Schinzel's hypothesis H implies that there exist arbitrarily large positive integers $n$ such that the numbers $P n+a_i$ and $(P n + a_i + 1)/(2P_i)$ are all prime, and such that $P n+j$ is not prime for $j \in J$. This makes the numbers $p_i:=P n + a_i$ consecutive primes such that $(p_i+1)/2 = P_i r_i$ for some prime $r_i$. If $n$ is sufficiently large, then these primes $r_i$ are all distinct and larger than all of the $q_S$. So the greatest common factor of $(p_i+1)/2$ and $(p_j+1)/2$ for $i \ne j$ equals $1$ if there is an edge between $i$ and $j$, and $q_{\{i,j\}}$ otherwise. $\square$
Remark: Given how little is known about consecutive primes, it seems unlikely that the conjecture can be proved unconditionally. But at least now we can be confident that it's true!