This is true (for large $m$ and $n$) under ABC plus the assumption that there is a prime in $[x,x+x^{1/2-\delta}]$ for some positive $\delta$ (which is widely believed, but beyond RH).  To see this, suppose $m <n$ 
and that they have the same radical $r$.  Write $m=gM$ and $n=gN$ where $g$ is the gcd of $m$ and $n$, so that $M$ and $N$ are coprime.  Applying the ABC conjecture to $M + (N-M) = N$, we conclude that 
$$ 
(N-M) r \ge N^{1-\epsilon},
$$ 
so that 
$$ 
n-m \ge n^{1-\epsilon}/r. 
$$ 
On the other hand, clearly $n-m$ is also $\ge r$ (since it is divisible by $r$).  It follows that $n-m \ge n^{1/2-\epsilon}$, and the assumption that there are primes in short intervals finishes the (conditional) proof.  

The problem is likely very hard, as fedja's observation in the comments already shows.  There is a conjecture of Hall that $|x^3-y^2| \gg x^{1/2-\epsilon}$ which is wide open.  The best results that are known here (going back to Baker's method) are of the flavor $|x^3-y^2| \gg (\log x)^C$ for some $C$.  If $x^3-y^2$ does get as small as in the Baker result, then take $n=x^4y$ and $m=xy^3$, which clearly have the same radical and then $|n-m|$ is of size essentially $n^{5/11}$.  In other words, either you have to improve work towards Hall's conjecture, or work towards gaps between primes!

**Added**  Thanks to Pasten's comment, I learned that this problem is already in the literature and is known as Dressler's conjecture.  The conditional proof above is recorded in [work of Cochrane and Dressler][1] who give more information on the conjecture.

[1]: http://www.ams.org/journals/mcom/1999-68-225/S0025-5718-99-01024-8/S0025-5718-99-01024-8.pdf