If we consider divisors to mean prime divisor, then this holds for most short intervals (less than length 6) and fails for some long intervals. 120 and 125 and 126 are smooth numbers between the same two consecutive primes; 24 and 27 are two others. In general, if you have an interval of 2k+1 consecutive composite numbers which contains a number whose prime factors are all at most k, this will spoil your conjecture.
Although it is not guaranteed to find all such examples, algorithm S from other MathOverflow questions (243490,248042) can be modified to find some other examples (such as 30600,30625).
If we are talking arbitrary divisors, then every number n has a divisor of n. If we are talking arbitrary proper divisors, then it is likely to be true as most numbers n have a divisor larger than the observed prime gap (larger than (log n)^2) and most n have a divisor larger than the asymptotic upper bound for prime gaps (O(n^(21/40)). This would mean (as observed in a comment above) that a spoiler would have to be a number with two equal or nearly equal nontrivial prime divisors, placed in the middle of an unexpectedly large prime gap.
Gerhard "It's Also About Smooth Numbers" Paseman, 2019.03.06.