The answer to your main question, about the Kullback--Leibler divergence, is no. Indeed, let the vectors $p=(1-s,s)$, $q=(1-t,t)$, and $r=p$ represent the corresponding probability distributions on (say) the set $\{1,2\}$, where $s\downarrow0$ and $t:=e^{-1/s^2}$. Then
$$KL(p||q)=(1-s)\ln\frac{1-s}{1-t}+s\ln\frac st\sim\frac1s\to\infty,
$$
$$KL(q||r)=KL(q||p)=(1-t)\ln\frac{1-t}{1-s}+t\ln\frac ts\to0,
$$
but $KL(p||r)=KL(p||p)=0$ -- which is of course not large but as small as $KL$ can be.

This phenomenon is general and not peculiar to distributions of a two-point set; it will occur on any nontrivial probability space. E.g., let the vectors $p=(\frac{1-s}{n-1},\dots,\frac{1-s}{n-1},s)$ and $q=(\frac{1-t}{n-1},\dots,\frac{1-t}{n-1},t)$ represent the corresponding probability distributions on the set $\{1,\dots,n\}$, for any natural $n\ge2$, where $s$ and $t$ are as above. Then we shall still have $KL(p||q)\to\infty$ but $KL(q||p)\to0$.

As for your "if not" question, there are many metrics on the set of all probability distributions. One of them is the Hellinger distance given by
$$d_H(p,q)=\frac1{\sqrt2}\sqrt{\int(\sqrt p-\sqrt q)^2}.
$$

For any such metric $d$ and any probability distributions $p,q,r$, by the triangle inequality we have $d(p,r)\ge d(p,q)-d(q,r)$. So, if $d(p,q)>M$ and $d(q,r)<t$, then $d(p,r)>M-t$.