First of all, for the question to make sense we must have $f\in L^r(X,\mu)$ for $p\le r<\infty$. If, as Didier suggests in his comment, $f\in L^\infty(X,\mu)$, then $$ \frac{\|f^n\|_p}{\|f^n\|_q}=\left(\frac{\|f\|_{np}}{\|f\|_{nq}}\right)^n. $$ Since $\lim_{r\to\infty}\|f\|_r=\|f\|_\infty$, $$ \lim_{n\to\infty}\frac{\|f\|_{np}}{\|f\|_{nq}}=1. $$ But this does not determine the desired limit.
Consider $f(x)=\chi_A(x)$, the characteristic function of a measurable subset $A\subset X$ with $\mu(A)>0$. Then $$ \frac{\|f^n\|_p}{\|f^n\|_q}=(\mu(A))^{1/p-1/q}, $$ which can be any number in $(0,1]$. As another example, if $X=[0,1]$ with Lebesgue measure and $f(x)=x^\alpha$, $\alpha>0$, then $$ \frac{\|f^n\|_p}{\|f^n\|_q}=\frac{(n\,\alpha\,q+1)^{1/q}}{(n\,\alpha\,p+1)^{1/p}}\to0. $$ My conjecture that for bounded $f$, the limit is one if and only if $f$ is constant almost everywhere.