Is a stack machine with a forward read iterator Turing complete?
Your model is Turing-complete, unfortunately.

You can simulate a queue in your data structure using the following algorithm. It introduced 3 new stack symbols: $d, x, y$.

Enqueue(val) is just Push(val).

For Dequeue():


ReadBegin().
Count the number of anything else - number of $d$ in the whole stack (which should be always non-negative). Push $y$ or pop $x$ for every $d$, and push $x$ or pop $y$ for anything else. Always prefer pop to push. Finally there won't be any $y$ in the stack and the result will be the number of $x$ on the top of the stack.
ReadBegin().
While pTop is a $x$:

Repeat ReadNext() until it returned something other than $x$ and $d$.
Pop().

Push a $d$.
The last result of ReadNext() is returned as the result of Dequeue.


The proof is straightforward. Check the revision history for a more complicated version firstly reducing it to a two-way version.