(More editing for cleanliness)

The statement is false. I learned of this example from "James" at this blog post. If you take a nodal cubic curve (notably quasiprojective), there is a flat, unramified cover by an infinite connected chain of copies of P^1, each glued transversely to its successor at a point. This is not profinite. If I'm not mistaken, the etale fundamental group of the nodal cubic over a separably closed field (with a chosen basepoint) is $\mathbb{Z}$, not its profinite completion.

**Edit:** Regarding the correct definition of etale fundamental group: In SGA1 Exp 5, Grothendieck (and Mme. Raynaud?) build up axiomatics for the theory of the fundamental group using only profinite sets, and the group is defined following one peculiar claim. In the beginning of Exp 5 Section 7, there is the assertion that for any connected locally noetherian scheme $S$, and any geometric point $a: \ast \to S$, the functor that takes an etale cover $X \to S$ to the set of geometric points over $a$ (with the usual morphisms) lands in the category of finite sets. The example I gave above seems to contradict this, but if you look in Exp 1, you find that all of SGA1 is written under a definition of etale morphisms that assumes that they are finite type (which this example is not). Anyway, one reason why Pete Clark only sees profinite definitions for the etale fundamental group, is that people like to use finite type morphisms, while etale morphisms only have to be locally of finite presentation (according to EGA4, and Wikipedia I guess).

As for the question of infinite degree etale covering maps between locally finite type geometrically integral schemes, I don't think one exists, since (if I'm not mistaken) you automatically get an infinite degree algebraic extension of function fields, which is therefore infinitely generated. I'm having trouble thinking through the details of this, though.