I'll consider the problem of computing $x^\frac1{q}, \; q > 0$; as I've already mentioned in the comments, one can decompose any positive rational number as $m+\dfrac{p}{q}$, where $m,p$ are nonnegative integers, $q$ is a positive integer, and $p < q$. Thus for computing $x^{m+\frac{p}{q}}$, one could use binary exponentiation on $x^m$ and $\left(x^\frac1{q}\right)^p$ and multiply the results accordingly.
A.N. Khovanskiĭ, in his book on continued fractions, displays a continued fraction representation for the binomial function:
$$(1+z)^\alpha=1+\cfrac{2\alpha z}{2+(1-\alpha)z-\cfrac{(1-\alpha^2)z^2}{3(z+2)-\cfrac{(4-\alpha^2)z^2}{5(z+2)-\cfrac{(9-\alpha^2)z^2}{7(z+2)-\cdots}}}}$$
which converges for $|\arg(z+1)| < \pi$.
Letting $z=x-1$ and $\alpha=\dfrac1{q}$, one can then evaluate this continued fraction (with, say, Lentz-Thompson-Barnett) to generate a "seed" that can be subsequently polished with Newton-Raphson, Halley, or any of a number of iterations with high-order convergence. You'll have to experiment with how accurate a seed you need to start up the iteration, by picking a not-too-small tolerance when evaluating the continued fraction.
Here's some Mathematica code demonstrating what I've been saying earlier, for computing $\sqrt[3]{55}$:
With[{q = 3, t = 55, prec = 30},
y = N[2 + (1 - 1/q) (t - 1), prec];
c = y; d = 0; k = 1;
While[True,
u = (k^2 - q^-2) (t - 1)^2; v = (2 k + 1) (t + 1);
c = v - u/c; d = 1/(v - u d);
h = c*d; y *= h;
If[Abs[h - 1] <= 10^-4, Break[]];
k++];
FixedPoint[
Function[x, x ((1 + q) t - x^q (1 - q))/(x^q (1 + q) - (1 - q) t)],
1 + 2 (t - 1)/q/y]]
Here, I've arbitrarily chosen to stop when the continued fraction has already converged to $\approx 4$ digits, and then polished the result with Halley's method. The result here is good to $\approx 28$ digits. Again, you'll have to experiment on the accuracy versus expense of evaluating the "seed", as well as picking the appropriate iteration method for polishing the seed.