Are there any odd primes like this? Are there any odd primes $p, q, r$ such that 

$$(p-1)(q-1)(r-1)\mid pqr-1$$
No, there are not.  Roughly speaking, because you can't have $(p-1)(q-1)(r-1)=pqr-1$, you need a factor of at least two between the sides, but the $pqr$ on each side is too big compared to the things you subtract.  

To sketch a more rigorous argument, we can demand $p \le q \le r$  If $p=3$, we are asking that $2(q-1)(r-1)|3qr-1$ or $2qr-2q-2r+2|3qr-1$  It is clear the two sides cannot be equal, so we need the right at least double the left.  To be double, we would need $q+r \gt \frac {qr}2$, which cannot happen as $q,r$ are too large.  The same happens even more if $p \gt 3$
Pseudoprimes do just this.  They pass Fermat's little theorm for all non-divisors. but are composite.  They are rarer then primes, but there are some classical ones.  One happens before 1000, and three come to mind less than 2000.