$\newcommand\ep\varepsilon$Of course not. 

E.g., suppose that $P(X=0,Y=0)=t\ep$, $P(X=0,Y=1)=1/2$, $P(X=1,Y=0)=(1-t)\ep$, and $P(X=1,Y=1)=1/2-\ep$, where $t$ and $\ep$ are in the interval $(0,1)$. Then 
$$I(X;Y)\sim c_t\ep$$
as $\ep\downarrow0$, where $c_t:=\ln2+t\ln t+(1-t)\ln(1-t)$. 

However, here $P(Y=0|X=0)\sim2t\ep$ as $\ep\downarrow0$, which is not relatively close to $P(Y=0)=\ep$ if (say) $t=1/4$.