I am looking for a proof that a finitely generated hyperbolic groups is non-amenable [unless it is virtually cyclic] which is a "metric/combinatoric" as possible. 

Here are the two proofs I am aware of: 1) use the arguments similar to "the centraliser of an element of infinite order is virtually cyclic" to show there is actually a free group [unless the group is virtually cyclic], 2) construct a Floyd boundary (or other ideal boundary) and ping-pong a free group out of the action on the boundary.

To make things slightly more precise, for a [finite] generating set $S$ and for a finite set $F$, define $\partial_S F$ to be the set of edges between $F$ and its complement in the Cayley graph with respect to $S$. 

$\mathbf{Question:~}$ What is a reference for a "as direct as possible" proof that a finitely generated hyperbolic groups satisfy: given $S$ as above, there is a $K>0$ such that for all $F$ finite, $|\partial_S F| \geq K |F|$?

Both proofs above use quite heavily the algebraic side of the problem. And there is a minimal algebraic content to any proof of the above implication: it is easy to construct a non-Cayley graph counter-example to the above question (take any hyperbolic Cayley graph and add a half-line to it). 

[Edit: just ignore the following paragraph]
PS: If it helps, on could assume that the group is finitely presented (i.e. it is the $\pi_1$ of some compact manifold). In fact, I would be delighted to see that this extra assumption can really be put to use.