Can we efficiently factor if we are given a Pocklington certificate of one of the prime factors?
Can we efficiently factor n?


Sure can, at least with that formulation of Pocklington's theorem.  From the certificate, we can obtain values $a_i, p_i$ with $a_i^{p_i} - 1 \equiv 0 \pmod p$, (where $p$ is the prime factor that the certificate applies to).

So, if we compute $a_i^{p_i} - 1 \bmod n$ (where $n = pq$), we know that that is 0 modulo $p$, but is extremely unlikely to be 0 modulo $q$ (the other prime factor).  So, compute the gcd of that and $n$, and that's the factor $p$.

On the other hand, if you go to the formulation on Wikipedia, instead of the listed criteria 4, we have $a^{n-1} \equiv 1 \pmod n$.  Of course, knowing a value $a$ that meets that criteria leaks nothing, as (assuming $n$ is prime) any $a$ that's between 1 and $n-1$ will meet that criteria.

If you create a certificate based on that, then it may be possible to factor, but that would depend on how the prime was generated (e.g. if the large factor $r$ of $p$ was not that much smaller than $p$, then it is easy; if it is just slightly larger than $\sqrt{p}$, then you need to hope that $(p-1)/r$ happens to be smooth (which is unlikely).