**EDIT.** I am returning to this many years after the fact. The essence of the answer remains unchanged, but I am writing it now as a more detailed proof.

Suppose that we are given a surface $S$ with a countable "$K$-qc atlas", where the change-of-charts maps are $K$-qc. (As already mentioned in the comments, it is important that the $K$ is uniform across $S$. I am assuming countable topology - this is true for Riemann surfaces by Rado's theorem; it should be true for your "qc" surfaces also, but I will just make it part of the definition.)

More precisely, the surface $S$ is covered by open sets $(U_i)_{i=1}^{\infty}$, and there are homeomorphisms $\phi_j\colon U_i\to D_i\subset \newcommand{\C}{\mathbb{C}}\C$ such that the transition maps $\phi_j\circ \phi_k^{-1}$ (when and where defined) are all $K$-qc.

We may assume for simplicity that the closures of the U_i are closed topological discs, every D_i is an actual Euclidean disc, and that any $U_i$ intersects at most finitely many other $U_k$. We won't actually use these, but they may simplify the mental picture.

**CLAIM.** There exist open sets $W_j\subset\C$ and $K$-quasiconformal homeomorphisms $\theta_i\colon D_i\to W_i$ such that the charts
$$ \psi_i\colon D_i\to W_j; \quad \psi_i := \theta_i\circ \phi_i $$
define a Riemann surface structure. (I.e., the transition maps $\psi_i\circ \psi_k^{-1}$ are all holomorphic.)

**PROOF OF THE CLAIM.** We construct the maps $\theta_i$ inductively, with the additional property that the dilatation of $\theta_i$ is trivial on
$$ \phi_i\left(U_i\setminus \bigcup_{k=1}^{i-1} U_k\right).$$
For $i=1$, this means that $\theta_1$ is conformal; we set $\theta_1 = \operatorname{id}$ and $W_1=D_1$. Suppose $i>1$ and the maps are defined up to $i-1$ with the desired properties. We define a Beltrami differential $\mu_i$ on $D_i$ as follows: for points in $\phi_i(U_k)$ with $k<i$, it is the pullback of the standard complex structure under $\phi_i\circ \psi_k^{-1}$. At other points, it is the standard complex structure. By the induction hypothesis, $\psi_{k'}^{-1}\circ \psi_k$ is holomorphic for $k'\leq k<i$, which means that the differential is well-defined.

Moreover, the dilatation of the differential is bounded by $K$. Indeed, for any point $z\in D_i$, let $k$ be minimal such that $z\in \phi(U_k)$. If $k=i$, then the dilatation is $1$ at $z$ by definition. Otherwise, it is given by the pullback of the standard dilatation under $(\phi_i\circ \phi_k^{-1})\circ \theta_k^{-1}$. Now $\theta_k$ is holomorphic at $\phi_k(z)$ by our additional inductive hypothesis, and the change of charts map is $K$-qc by assumption. So the dilatation is indeed bounded by $K$.

Now we can solve the Beltrami equation and find $W_i$ and a quasiconformal map $\theta_i\colon D_i \to W_i$ such that $\mu_i$ is the pullback of the standard complex structure on $\theta_i$. (If $U_i$ is connected and simply-connected, we can let $W_i$ be a disc.) This means that $\psi_i\circ \psi_k^{-1}$ is holomorphic for $k<i$, as desired. This completes the inductive construction and the proof of the claim.

Observe that the map from our original "qc surface" to this new Riemann surface is, in charts, given by the $\theta_i$. So it is $K$-quasiconformal, and we have found the desired Riemann surface structure.

The argument works for any surface; I see no issue with extending it to surfaces with boundaries.

Note that, if we choose a different enumeration of our charts, then we will get a different Riemann surface structure. So this structure is not canonical, as mentioned in the original version of my answer.