Given reals $A,B,X$, let $A\le_{T/X}B$ iff $A\oplus X\le_TB\oplus X$. For each real $X$ we can define a version of the c.e. degrees over $X$: we look at the preorder on $X$-c.e. reals given by $\le_{T/X}$ (or, if preferred, take the corresponding partial order). Call this $\mathcal{R}_X$; equivalently, $\mathcal{R}_X$ could be given by the reals which are CEA $X$.

My question is the following:

Suppose $G$ is "sufficiently" Cohen generic. Is $\mathcal{R}_G\cong\mathcal{R}$?

Here $\mathcal{R}$ is the usual c.e. degrees. Incidentally, it's not hard to show that if $G,H$ are "sufficiently" Cohen generic then $\mathcal{R}_G\cong\mathcal{R}_H$. I vaguely recall a result (due to Shore?) that the answer is no, but I can't track it down.

If the answer is no, I'm curious how similar they are nonetheless. For example, do $\mathcal{R}_G$ and $\mathcal{R}$ have the same $\Pi_3$ theories?

Given reals $A,B,X$, let $A\le_{T/X}B$ iff $A\oplus X\le_TB\oplus X$. For each real $X$ we can define a version of the c.e. degrees over $X$: we look at the preorder on $X$-c.e. reals given by $\le_{T/X}$ (or, if preferred, take the corresponding partial order). Call this $\mathcal{R}_X$; equivalently, $\mathcal{R}_X$ could be given by the reals which are CEA $X$.

My question is the following:

Suppose $G$ is "sufficiently" Cohen generic. Is $\mathcal{R}_G\cong\mathcal{R}$?

Here $\mathcal{R}$ is the usual c.e. degrees. Incidentally, it's not hard to show that if $G,H$ are "sufficiently" Cohen generic then $\mathcal{R}_G\cong\mathcal{R}_H$. This is false. 

I vaguely recall a result (due to Shore?) that the answer is no, but I can't track it down.

If the answer is no, I'm curious how similar they are nonetheless. For example, do $\mathcal{R}_G$ and $\mathcal{R}$ have the same $\Pi_3$ theories?