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they as does, for example, the $r$-matrice of the Drinfeld double of a cocommutative Lie bialgebra. They also prove that $R=\exp(r)$ ($r:=r^{1,2}$) satisfies the Quantum Yang Baxter equation. Therefore, one get this way a representation of the (pure) braid group in $\exp(\mathfrak{tder}_n)\rtimes S_n$. The $r^{i,j}$ can be thought of as "arrow diagrams" and are related to the welded braid group.

Hence $F$ is a universal version of a Drinfeld twist for Quasi-Hopf algebra, which is able to "kill" the associator, and according to the theory it implies that the corresponding representations of the braid group are the same.

Edit 2: This was actually the situation for braids, let me add a few word about knots. It turns out that while usual braids embeds into welded one, this is far from being true at the level of knots. So the twist also intertwines between the restriction of the invariant for welded knots to usual knots, and the image of the Kontsevich integral in the space of arrow diagrams, but the resulting invariant is much weaker.

Welded braids can be identified with the group of basis conjugating automorphisms of a free group, and the map it gets from usual braids is nothing but the Artin representation. This representation is faithfull, but its extension to string links (and in particular to long knots) has a huge kernel. At the level of diagrams, the algebra corresponding to knots is free commutative with two generators in degree one, and one generator in each degree greater than one. Hence it has a rather simple structure, while its analog for usual knots is very complicated.

In fact, one of the main claim of Bar Natan's paper is that the universal invariant for welded knots is roughly the Alexander polynomial.

However, let me mention that usual knots does embed into virtual one, and that the above story is an important step towards something similar in the virtual case.

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As you point out, the relation between associators and the quasi-triangular structure of $U_q(\mathfrak g)$ (and the related tangles invariants) exists "only" at the Lie algebraic level, not (not yet) at the universal one. Roughly speaking, this is because the twisting which absorb the associator is not $\mathfrak g$-invariant, ie modifies the coalgebra structure, which a priori doesn't make sense at the level of chord diagrams.

So far I know, there is no combinatorial construction of a universal finite type invariant which can avoid associators. But of course it's something people are looking for.

But it turns out that the theory of quantum $R$-matrices is more related to the theory of virtual knotted objects (see this answer of Greg Kuperberg). This is more or less "Bar Natan's dream" that a universal finite type invariant for virtual knotted objects should corresponds somehow to Etingof--Kazhdan quantization of Lie bialgebras.

There is also a baby version of this, which is Alekseev-Enriquez-Torrossian solution of the Kashiwara Vergne conjecture based on associators. It turns out that they constructs a kind of universal twist which can "kill" the associator, and a kind of universal solution of the quantum Yang-Baxter equation, living in a bigger algebra than the algebra of horizontal chords diagrams. According again to Bar Natan, this corresponds more or less to a universal finite type invariant for "wedded welded knots". See: http://www.math.toronto.edu/drorbn/papers/WKO/

You may also find this paper interesting : Towards a Diagrammatic Analogue of the Reshetikhin-Turaev Link Invariants

Edit: Some details about the relation between the Alekseev-(Enriquez)-Torossian construction and Vassiliev invariants.

They start from the Lie algebra $\mathfrak{tder}_n$ of "tangential derivations" of the free Lie algebra $\mathfrak{f}_n$ on $n$ generators, that is the Lie algebra of endomorphism sending each generator $x_i$ to $[x_i,a_i]$ for some $a_i \in \mathfrak f_n$.

Let $r^{i,j}$ be the element mapping $x_i$ to $[x_i,x_j]$, and $x_k$ to 0 for $k\neq i$. Then it leads to a solution of the "classical Yang-Baxter equation whose second leg lives in a commutative subalgebra", i.e.: $$[r^{1,3},r^{2,3}]=0$$ and $$[r^{1,2},r^{1,3}+r^{2,3}]=0$$ they also prove that $R=\exp(r)$ ($r:=r^{1,2}$) satisfies the Quantum Yang Baxter equation. Therefore, one get this way a representation of the (pure) braid group in $\exp(\mathfrak{tder}_n)\rtimes S_n$. The $r^{i,j}$ can be thought of as "arrow diagrams" and are related to the welded braid group.

On the other hand, exactly like in the usual theory of the Yang-Baxter equation, we have that $t^{i,j}=r^{i,j}+r^{j,i}$ satisfies the infinitesimal braid relations, also called 4t relation for horizontal chord diagrams. We thus get an injective morphism from the algebra of Horizontal chord diagram into $U(\mathfrak{tder}_n)$. Therefore we can put an associator and get another representation of the braid group in $\exp(\mathfrak{tder}_n)\rtimes S_n$ which is precisely the "Kontsevich integral for braids".

Then the main result of [AT] is the following identity: let $\Phi$ be (the image in $\exp(\mathfrak{tder}_3)$ of) an associator, then there exists some $F \in \exp(\mathfrak{tder}_2)$ such that $$F^{2,3}F^{1,23}=\Phi F^{1,2} F^{12,3}$$

where the indices correspond to some maps $\mathfrak{tder}_2 \rightarrow \mathfrak{tder}_3$ modelled on the coproduct of an envelopping algebra. They also show that:

$$R=F e^{t/2} (F^{2,1})^{-1}$$

Hence $F$ is a universal version of a Drinfeld twist for Quasi-Hopf algebra, which is able to "kill" the associator, and according to the theory it implies that the corresponding representations of the braid group are the same.

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Edit: Some details about the relation between the Alekseev-(Enriquez)-Torossian construction and Vassiliev invariants.

They start from the Lie algebra $\mathfrak{tder}_n$ of "tangential derivations" of the free Lie algebra $\mathfrak{f}_n$ on $n$ generators, that is the Lie algebra of endomorphism sending each generator $x_i$ to $[x_i,a_i]$ for some $a_i \in \mathfrak f_n$.

Let $r^{i,j}$ be the element mapping $x_i$ to $[x_i,x_j]$, and $x_k$ to 0 for $k\neq i$. Then it leads to a solution of the "classical Yang-Baxter equation whose second leg lives in a commutative subalgebra", i.e.:they also prove that $R=\exp(r)$ ($r:=r^{1,2}$) satisfies the Quantum Yang Baxter equation. Therefore, one get this way a representation of the (pure) braid group in $\exp(\mathfrak{tder}_n)\rtimes S_n$. The $r^{i,j}$ can be thought of as "arrow diagrams" and are related to the welded braid group.

On the other hand, exactly like in the usual theory of the Yang-Baxter equation, we have that $t^{i,j}=r^{i,j}+r^{j,i}$ satisfies the infinitesimal braid relations, also called 4t relation for horizontal chord diagrams. We thus get an injective morphism from the algebra of Horizontal chord diagram into $U(\mathfrak{tder}_n)$. Therefore we can put an associator and get another representation of the braid group in $\exp(\mathfrak{tder}_n)\rtimes S_n$ which is precisely the "Kontsevich integral for braids".

Then the main result of [AT] is the following identity: let $\Phi$ be (the image in $\exp(\mathfrak{tder}_3)$ of) an associator, then there exists some $F \in \exp(\mathfrak{tder}_2)$ such that$$F^{2,3}F^{1,23}=\Phi F^{1,2} F^{12,3}$$

where the indices correspond to some maps $\mathfrak{tder}_2 \rightarrow \mathfrak{tder}_3$ modelled on the coproduct of an envelopping algebra. They also show that:

$$R=F e^{t/2} (F^{2,1})^{-1}$$

Hence $F$ is a universal version of a Drinfeld twist for Quasi-Hopf algebra, which is able to "kill" the associator, and according to the theory it implies that the corresponding representations of the braid group are the same.

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