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).

First, I claim any transformation that is $\delta$-close to the identity for $\delta<\frac 12$ 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 $A_j=\{x\colon x=\min(x,Tx,\ldots,G^{j-1}x)\}$. Let $B_j=\bigcup_{k<j;\text{ $k$ even}}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 $\int |\mathbf 1_B\circ I-\mathbf 1_B\circ G|>\frac 12$. 

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 $T$.