$\newcommand\g{\mathfrak{g}}$The answer to your question "is there a procedure that takes $\g$ as input, produces $U_q(\g)$ as output, and doesn't involve the choice of a Cartan subalgebra of $\g$?" is No. Not if you want it "canonical" in any sense. (Of course, if I wanted to cheat my way to a "yes," I could make choices that are equivalent to choosing a Cartan but not stated as such — I would construct for you a certain canonical homogeneous space, and then ask you to pick a point in it, ....)

The problem is that the automorphism group of $\g$ does not lift to $U_q(\g)$. Recall that the inner automorphism group is precisely the simplest group $G$ integrating $\g$ (take any connected group integrating $\g$ and mod out by its center). On the other hand, the inner automorphism "group" of $U_q(\g)$ is $U_q(\g)$ (or "$\operatorname{spec}(\operatorname{Fun}_q(G))$", depending on your point of view) itself. This certainly deforms the automorphisms. But there is not a procedure like you ask for, because you have to break some symmetry.

Here's a way to say this correctly: In a precise sense $U_q(\g)$ degenerates to $U(\g)$ as $q\to 1$, and this is part of the structure that I take you to mean when you write "$U_q(\g)$". In this degeneration, you can also study $\frac{\partial}{\partial q}(\dots)$ at $q=1$. In particular, looking at $\frac{\partial}{\partial q}\bigr|_{q=1}$ of the comultiplication on $U_q(\g)$ recovers the Lie cobracket on $\g$. But the Lie cobracket knows the Cartan subalgebra: it is precisely the kernel of the Lie cobracket. So $\operatorname{Aut}(\g)$ cannot lift to $U_q(\g)$ compatibly with all of this structure.

What does exist without choosing a Cartan subalgebra is the braided category of representations of $U_q(\g)$, although I would have to think a moment to recall how to write it down explicitly. (In the asymptotic limit $q = e^\hbar$ with formal $\hbar$, I do know how to write down $\operatorname{Rep}(U_{e^\hbar}\g)$ explicitly for any choice of Drinfel'd associator.) In particular, as a braided monoidal category, this category does have an action by $\operatorname{Aut}(\g)$.

But the category is strictly less data than the Hopf algebra. Namely, "Tannakian reconstruction" is the statement that $U_q(\g)$ is the Hopf algebra of endomorphisms of a certain braided coalgebra in $\operatorname{Rep}(U_{q}\g)$ (or it is the Hopf dual of the Hopf algebra of co-endomorphisms of a certain braided algebra in the category of Ind-objects in $\operatorname{Rep}(U_{q}\g)$, if for you representations are finite-dimensional), and you cannot choose this coalgebra canonically. This coalgebra is unique up to isomorphism, but certainly not up to unique isomorphisms (or else $U_q(\g)$ would be trivial). The failure of this coalgebra to exists up to canonical isomorphism is essentially the same problem as above.

In a precise way, this is failure of there to exist a canonical isomorphisms between different choices of the coalgebra is analogous of the failure of the "fundamental group" of a topological space to be an honest group. Recall that a *pointed* topological space has a fundamental group, which is a group determined up to *canonical* isomorphism. For comparison, a non-pointed but path-connected topological space has a group assigned to each point, and these are non-canonically isomorphic. Thus a non-pointed path-connected topological space has a "fundamental group up to conjugation," also known as a *connected groupoid*.

What this should all mean, although I'm not going to try to write down the correct definition, is that there does exist canonically associated to $\g$ a "Hopf algebroid" which is noncanonically equivalent as a Hopf algebroid to any choice of $U_q(\g)$. Or, at least, I'm confident of everything in my answer in the $\hbar\to 0$ asymptotics, and have thought less about the finite-$q$ case, and so I'm generalizing intuition from that setting, but I think it's all correct.