Question: Let $G$ be an ample $\mathbb{Q}$-divisor on a smooth variety $X$. Let $D$ be a $\mathbb{Q}$-divisor linearly equivalent to $G$. Let $f: Y\to X$ be a common log resolution of $G$ and $D$. We define the multiplier ideal of a divisor $G$ as $I(G)=f\_\*O\_Y(K\_{Y/X}-[f^\*G])$. Are the multiplier ideals $I(G)$ and $I((1-t)G+tD)$ the same for sufficiently small $t>0$?

Answer: 
I don't think so. Take for example $X=\mathbb{P}^2$, and $G$ and $D$ to be distinct lines.
then $I(G)=\mathcal{O}\_X(-G)$ while $I((1-t)g+tD)=\mathcal{O}\_X$ for every small $t>0$.
Maybe you might want to look at the multiplier ideal associated to the linear series $|G|$.
However, I should point out that one always has the multiplier ideal containment
$$I((1−t)G+tD) \supseteq I(G)$$ for $t > 0$ sufficiently small.

In particular, the singularities of $(X, (1-t)G + tD)$ can only be epsilon more severe than the singularities of the pair $(X, G)$, thus the multiplier ideal won't get smaller, at least for sufficiently small $t$. Furthermore, they might even be better (in other words, less severe) as Gianni's example shows.