I'm assuming that by $\delta$-close, you mean $\int |g\circ T-g\circ G|\,d\mu<\delta\|g\|_\infty$. Without the absolute values, everything would be $\delta$-close. 

So the answer is no. Here is a proof. The constants are not optimized (at all). I should say (to make it clear where this comes from) is that this is kind of obvious if you're familiar with Rokhlin's lemma. Rokhlin's lemma is the surprising (but simple) result that if $T$ is an ergodic (or in fact if the set of periodic points has measure 0), then for every integer $n$ and every $\epsilon>0$, there exists a set $B$ such that $B$, $T^{-1}B$, ... $T^{-(n-1)}B$ are disjoint and are each of measure $(1-\epsilon)/n$. The remainder is called the "error set", $E$. The picture is that the "levels of the tower", $T^{-(n-1)}B$, ..., $B$ are arranged one above the other and $T$ just moves a point from one level to the next one up. Once it gets to the top, a point may move to the bottom; or it may move to $E$. From $E$ it may move to $E$ or to the bottom of the tower. 

First, I claim any transformation that is $\delta$-close to the identity for $\delta<\frac 14$ is equal to the identity on a set of measure at least $\frac14$. 

To prove this suppose that $G$ is a transformation that is equal to the identity on a set of measure less than $\frac 14$. Then let $X_j$ be the (measurable) subset of $X$ consisting of points of least period $j$. Then there exists a measurable subset $A_j$ of $X_j$ such that $X_j=A_j\cup G^{-1}A_j\cup \ldots\cup G^{-{j-1}}A_j$ (for instance $A_j=\{x\colon x=\min(x,Gx,\ldots,G^{j-1}x)\}$. Let $B_j=\bigcup_{k<j;\text{ $k$ odd}}G^{-k}A_j$. 

Let $X_\infty$ be the remaining set of non-periodic points. Assuming $X_\infty$ has positive measure, then there exists (by Rokhlin's theorem) a subset $B_\infty$ of $X_\infty$ of measure $\frac 13\mu(X_\infty)$ such that $B_\infty\cap G^{-1}B_\infty=\emptyset$. 

Finally set $B=(\bigcup_{j=2}^\infty B_j)\cup B_\infty$. Notice that if $x\in B$, then $Gx\not\in B$. Also $\mu(B)\ge \frac 13(\mu(X)-\mu(X_1))>\frac 14$. Hence $\int |\mathbf 1_B\circ I-\mathbf 1_B\circ G|>\frac 14$. 

Secondly, I claim that if $T$ is ergodic, and $H$ is the identity on a set of measure at least $\frac 14$, then $T$ and $H$ are not $\frac 7{44}$-close.

Again, by Rokhlin's lemma, let $B$ be a set of measure $\frac 1{11}$ such that $B$, $T^{-1}B,\ldots,T^{-9}B$ are disjoint and let $f$ be the indicator function of $B\cup T^{-2}B\cup\ldots\cup T^{-8}B$. Then $\{x\colon f(x)\ne f(T(x))\}$ has measure at least $\frac {10}{11}$. On the other hand, 
$\{x\colon f(x)\ne f(H(x))\}\le \frac 34$ since $H$ is the identity on a set of measure at least $\frac 14$. 
On the symmetric difference of these sets, $|f(H(x))-f(T(x))|=1$. That is, on a set of measure at least $\frac 7{44}$. 

But if $G$ is $\frac 1{10}$-close to the identity and $G^n$ is $\frac1{10}$-close to $T$, then by the first part, $G$ is the identity on a set of measure at least $\frac 14$, and so $G^n$ is also the identity on a set of measure at least $\frac 14$. But then by the second fact, $G^n$ is not $\frac7{44}$-close to $T$.