The first counter-example is for $p=17$. The interval of length $40$ starting at $87890$ only yields three integers with largest prime factor greater than $17$: the primes $87917,87931$ and also $87929=23\cdot 3823$. So we can use the inteval of length $38$ starting at $87890$ or at $87891$.

Since we need only check numbers which are $1,7,11,13,17,19,23,29 \mod 30$ we can verify this by noting that 
$87893=13\cdot6761,\, 87899=7\cdot29\cdot433,\, 87907=17\cdot 5171$ $87913=17\cdot 19 \cdot 661,\, 87919=13\cdot 6763$ and $87923=11 \cdot 7993.$

There are essentially 6 ways to get a run of length $39$ (or 3 up to reflection): One can start at $87890,177980,182342,328130,332492$ or $422582 \mod 510510$. 

There are not any counter-examples for $p=19$. The 43 consecutive integers from $259878$ to $259920$ all have a divisor $19$ or less with the two exceptions $259891$ and $259907$ but there are no cases longer than that. There are essentially 6 ways (or 3 up to reflection) to get a run that long.