An easy answer along traditional lines is available iff the measure has "bounded variation" in a suitable sense.

To undestand this, first note that integration with values in Banach algebras (which are rings, but not compact) contains the integration of scalar valued functions with respect to vector (Banach) valued measures (embed the Banach space in a ring of continuous functions and identify scalars with constant functions). Then check the books with the theory of (strong) integration with respect to vector valued measures (even some book about Banach lattices have this). Or directly note that to have a Cauchy condition on sums of terms $f(x_I)m(I)$ (where the clopens $I$ form a finite partition of $X$) the easy way to control by means of the triangle inequality the sum of the $[f(x_I)-f(y_I)]m(I)$ when $f$ is uniformly continuous is to suppose that the sum of the $\|m(I)\|$ is bounded by a constant independent from the partition: i.e. $m$ has bounded variation.

I think that the problem is unaivoidable even if you try the "weak" theory of integration. [More on this below].

So suppose that you have three (abelian, additive) monoids with a uniform structure that makes sum uniformly continuous (examples: topological groups; compact monoids with continuous operation). Say $A$ where measures take their values, $B$ where functions take their values, and $C$ (taken to be complete; if not, complete it and the other axioms are not lost) where intergrals take their values; cleary one needs a locally uniformly continuous biadditive multiplication from $A\times B$ to $C$. [Perhaps one should weaken all axioms to separate local uniform continuity of all binary operations; except for the completion process of $C$ the remaining things might work].

Continuous functions on the compact $X$ are uniformly continuous, and by 0-dimensionality they are exactly the uniform limits of step functions. Riemann sum of terms $f(x_I)m(I)$ always satisfy a Cauchy condition (for all continuos $f$) iff the fixed finitely additive measure $m$ satisfies a suitable "bounded variation" condition.
$C$ to the reals (or other structures $S$ where $S$-valued integrals are
When there are sufficiently many continuous homomorphisms $\phi$ from already defined), one could also use the "weak" integral: $\int fdm=c$ iff for each homomorphism $\phi$ the Riemann sums of the terms $\phi(f(x_I)m(I))$ converge to $\phi(c)$. The corresponding notion of "bounded variation" for $m$ might be weakened, but not trivialized (infact, for complex valued measures and functions nothing changes).

Potential bibliography: perhaps the memoir AMS of McShane (1967 or so) where he defines very general integrals (including Lebesgue integral with respect to a Radon measure) as limits of Riemann's sums (in the style of Henstock - Kurzweil). The primary objective of such memoir is not absolute generality in the range of functions, measures and integrals, but if I remember correctly there is some substance also in this direction. In years around that date there was empasis on generailty, so it might be also useful the bibliography of Henstock's book "general theory of integration" (1991 or so). Another advantage of looking at Henstock - style integration is that it is useful also in cases where the measure is not of bounded variation (and the integral is nonabsolutely convergent): for example, integration with respect to Wiener's measure on the space of continuous functions is a ordinary integration with respect to a positive measure, but when physicists need the analogue where the gaussian terms exp$(-x^2)$ are replaced by exp$(ix^2)$ (Feynmann's path integrals), then one has lost (not $\sigma$-additivity on a non-$\sigma$-algebra of sets, but) bounded variation (Fresnel's integrals are nonabsolutely convergent) and ordinary measure theory does not work, but Henstok's general theory works (directly, without looking at $i$ as limit of $i-\epsilon$ and using the absolutely convergent integrals for $i-\epsilon$).