I'm not sure what you mean by a random integer $n$, but would you agree that the probability that a random squarefree integer be divisible by $55$ is nonzero? For if $55\mid n$ then $5\mid\gcd(n,\phi(n))$.
Added In Derek's new notation it's well-known that $q(N)\sim 6N/\pi^2$. This constant arises via $$\frac6{\pi^2}=\prod_p \left(1-\frac1{p^2}\right).$$ Sticking to the example of the number $55$, if $g_{55}(n)$ is the number of squarefree numbers up to $n$ that are divisible by $55$ then $g_{55}(n)=h_{55}(n/55)$ where $h_{55}(n)$ is the number of squarefree numbers up to $n$ that are not divisible by $55$. But $h_{55}(n)\sim\alpha n$ where $\alpha$ is the same Euler product with the $p=5$ and $p=11$ terms dropped: $$\alpha=\frac6{\pi^2}\frac{5^2\times 11^2}{24\times 120}.$$ Thus $$\frac{g_{55}(n)}{q(n)}\to\frac{55}{24\times 120}=\frac{11}{576}$$ as $n\to\infty$, but of course $f(n)/q(n)\ge g_{55}(n)/q(n)$.
Of course one can perform this argument with other numbers in place of $55$.