In fact, it is a well-known observation in cryptography that just knowing a multiple of $\phi(n)$ helps greatly in factoring it regardless of the number of prime factors (if there's only one prime dividing $n$ then this has already been covered).
Suppose $m$ is a multiple of $\phi(n)$. Then if you factor out enough powers of $2$, there must exist a divisor $t = m/(2^r)$ such that $\lambda(n) \mid 2t$ but $\lambda(n) \nmid t$. (Here, $\lambda(n)$ is the Carmichael function, which is even unless $n$ is trivially small.)
It will then happen that for some bases $b$, $b$ will be a quadratic residue for some prime $p \mid n$ but not for a different $q \mid n$. In this case, taking $(b^t-1,n)$ will produce a non-trivial factor of $n$.
One simply has to randomly try different values of $b$ (the expectation is finite), as well as different choices of $t$ (there are at most $\log(n)$ possibilities, and one can use successive squaring to efficiently cover them).
To get prime factors of $n$, just repeat the process (we still have a multiple of $\phi$ for both of those factors).