Information-theoretic Diffie-Hellman
No, there is no information-theoretic analog that is secure against computationally-unbounded adversaries.

To form an analog, we'd need an injection $\varphi$ that maps $x$ in fine representation to $x$ in coarse representation.  But then Diffie-Hellman involves Alice sending $\varphi(x)$ publicly, and Bob sending $\varphi(y)$ publicly.  An eavesdropper can see $\varphi(x),\varphi(y)$.  Since $\varphi$ is an injection, information-theoretically this reveals $x,y$, which is enough to reveal the negotiated key.

What if we consider a function $\varphi$ that is non-injective?  This doesn't help.  Consider the equivalence relation $\sim$ where $x \sim x'$ if $\varphi(x)=\varphi(x')$, and let $[x]$ be the equivalence class of $x$.  Then for the scheme to work, the negotiated secret has to depend only on $\varphi(x),y$, i.e., only on $[x],y$ (since Bob has to be able to compute it).  Similarly, the negotiated secret has to depend only on $x,[y]$.  It follows that the negotiated secret depends only on $[x],[y]$.  Why?  Let $f(x,y)$ denote the secret negotiated if Alice uses fine value $x$ and Bob uses fine value $y$.  Suppose $x \sim x'$ and $y \sim y'$.  Then since $[x]=[x']$, it follows that $f(x,y)=f(x',y)$ (since $f(x,y)$ depends only on $[x],y$).  Also since $[y]=[y']$, it follows that $f(x',y)=f(x',y')$ (since $f(x,y)$ depends only on $x,[y]$).  Therefore $f(x,y)=f(x',y')$ whenever $x=x'$ and $y=y'$.  In other words, the final negotiated secret depends only on $[x],[y]$.  However, the eavesdropper can observe $\varphi(x),\varphi(y)$, from which $[x],[y]$ are uniquely determined and thus the negotiated secret is uniquely determined and thus information-theoretically cannot be hidden from the eavesdropper.

Public-key exchange can only be secure computationally -- you can't achieve information-theoretically secure public-key exchange.