Patrick,
Yes it does travel farther before reaching natural roll, but it's not all that obvious why. Imagine that the friction force is constant as the cueball goes from zero up to natural roll speed. (This isn't a necessary assumption, and isn't exactly true, particularly right after impact (bounce), but it's very nearly true and simplifies things.) If you plotted the cueball's ever increasing speed against time, it would be a straight line sloping upwards. Let's say at time T it reaches natural roll at speed V'. So how far did it travel?
I think it's somewhat obvious, although you might require a more rigorous argument to be convinced, that its average speed over this period is V'/2. It's V'/2 because it went from zero to V' in a linear fashion. Call it Vav. The distance traveled, D, during a time T, by anything, no matter how complicated its speed might vary over time, is Vav*T (the asterisk denotes simple multiplication). Since Vav=V'/2, and since the natural roll speed V' is independent of the magnitude of cloth friction (i.e., it doesn't matter how quickly it achieves natural roll given the same initial spin), Vav is also independent of this. And since D=Vav*T, the distance D depends only on T, and is greater as T is longer.
Sorry if I've made things worse, but it's not all that easy (for me) to provide a more straightforward argument when you have a continuously varying speed. That's why calculus was invented, although it isn't too much of a leap to go there (mainly, divide the time interval into very small increments of dt and add up all the contributions of Vdt as V varies with t). Maybe someone (smarter) can conjure up a more intuitive description if the above doesn't persuade.
Another way of looking at it is to consider the distance if the speed was absolutely constant throughout and equal to V' right from the start. The plot would have speed as a straight line parallel to the time axis at height V'. The distance traveled would be V'*T. This is the "area" under a rectangle with sides of length V' and T. "Area" doesn't mean a spacial area in the normal sense, of course, but a abstract representation of the distance as such. When the speed ramps up linearly to V', we have a triangle instead of a rectangle,. The area of the triangle is (1/2)V'*T as with any triangle (1/2 base X height). If you accept this "area" as a legitimate representation of distance traveled, and that V' only depends on the retained spin after CB-OB impact, you reach the same conclusion.
Not sure if any of that helps, but I'm honestly not trying to obscure with the damned math. Once you're comfortable with it, it tells you things that are sometimes very hard or impossible to see otherwise.
Jim