What follows is a hands-on explanation (not a complete proof!) of why the Tannakian group $$G_{\textrm{dR}}(AM(\mathbb{Q}))$$ of the category of Artin motives is a nontrivial inner form of $$G_{\textrm{B}} (AM(\mathbb{Q})) = \underline{\textrm{Gal}}_\mathbb{Q}.$$
Ironically, the source for this answer is actually the original asker, who three years later published an interesting paper that clarifies this point: https://arxiv.org/abs/1706.06573. All I've done is translate a sliver of his work into very concrete language for the benefit of posterity.
I'm actually going to work with a single 0-dimensional motive $X = \mathfrak{h}(\textrm{Spec } k)$ for a finite Galois extension $k/\mathbb{Q}$. This will produce a Tannakian group $G_{X, \bullet}$ associated to the subcategory of motives $\langle X \rangle$ (motives generated from $X$ by tensor powers, duals, and direct sums). $G_{X, \bullet}$ will be a quotient of the full Galois group for 0-dimensional motives; it should be clear how to fit the different $G_{X, \bullet}$ together compatibly to build the full groups $G_{\bullet}(AM(\mathbb{Q}))$. (It's just like building the full absolute Galois group out of its action on finite Galois extensions!)
For completeness, let's first see why the Betti realization is the Galois group viewed as a constant group scheme. According to the Tannakian formalism, a $K$-point of the algebraic group $G_{X,\textrm{B}}$ is a system of $K$-linear automorphisms $$T_Y: H_B(Y, \mathbb{Q}) \otimes K \to H_B(Y, \mathbb{Q}) \otimes K$$ for each 0-dimensional motive $Y \in \langle X\rangle$ such that the following diagram commutes for all maps of motives $\varphi: Y \to Y'$:
$\require{AMScd}$
\begin{CD}
{H_B(Y, \mathbb{Q}) \otimes K} @> {T_Y}>> H_B(X, \mathbb{Q}) \otimes K\\
@V{H_B(\varphi)} VV @V{H_B(\varphi)}VV\\
{H_B(Y', \mathbb{Q}) \otimes K} @> {T_{Y'}}>> H_B(Y', \mathbb{Q}) \otimes K\\
\end{CD}
The system $T_Y$ is also required to be compatible with tensors and duals.
Let's first think about our generating motive $X$. We know that $H_B(X, \mathbb{Q})$ is the dual of the vector space spanned by $\textrm{Spec }\mathbb{C}$, the set of embeddings $k \to \mathbb{C}$. Maps of Artin motives $X \to X$ all happen to be (linear combinations of maps) of the form $g: k \to k$, where $g$ is an element of $G = \textrm{Gal}(k/\mathbb{Q})$. Since $g$ acts on an embedding $k \to \mathbb{C}$ by pullback (precomposition), the $G$ action on $\textrm{Spec }k (\mathbb{C})$ is the left regular representation (left multiplication on $\mathbb{Q}[G]$); then the action on $H_B(X, \mathbb{Q})$ is the dual, also known as the right regular representation (multiplication on the right by $g^{-1}$). In other words, in the special case of a morphism $g: X \to X$ induced by a Galois element $g: k \to k$, our commutative diagram becomes
\begin{CD}
{\mathbb{Q}[G] \otimes K = K[G]} @> {T_X}>> K[G]\\
@V R_g VV @V R_g VV\\
K[G] @> {T_X}>> K[G]\\
\end{CD}
Just as in the Tannakian formalism for a finite group $G$, we need the further constraint on $T_X$ that comes from morphisms between $X \times X$ and $X$. Consider the multiplication morphism $k\otimes k \to k$; it induces a map $f: H_B(X) \otimes H_B(X) \to H_B(X)$. Under the isomorphism $H_B(X) \cong \mathbb{Q}[G]$, this map is $$g \otimes h \mapsto \begin{cases} 0, & \textrm{if } g\neq h \\ g, & \textrm{if } g = h. \end{cases}$$
(To check this, recall that we've identified $g \in \mathbb{Q}[G]$ with the dual basis element to $\iota \circ g$ in $\textrm{Spec }\mathbb{C}$. So $g\otimes h$ is dual to the map $k \otimes k \to \mathbb{C}$ that embeds the first factor by $\iota \circ g$ and the second by $\iota \circ h$, then multiplies them in $\mathbb{C}$. To evaluate $f(g \otimes h)$ on an embedding $\iota \circ g': k \to \mathbb{C}$, we pull back to $k\otimes k \to k \to \mathbb{C}$ and check whether we got the embedding which acts as $\iota \circ g$ on the first factor and $\iota \circ h$ on the second. We will get 0 unless $g = h = g'$.)
(Note: this map $f$ is actually the same as the cup product map, because $k\otimes k \to k$ by multiplication is the diagonal morphism.)
Now our map $T_X$ is supposed to make the following diagram commute:
\begin{CD}
K[G] \otimes K[G] @> {T_X \otimes T_X}>> K[G] \otimes K[G] \\
@V f VV @V f VV\\
K[G] @> {T_X}>> K[G]\\
\end{CD}
(Note that $f$ and $T_X$ are extended linearly to $H_B(X) \otimes K$. To reiterate, this is because we are looking for $K$-points of the motivic Galois group.)
At this stage, we're basically reduced to some annoying calculations. It turns out that there is only one way to make these diagrams commute: $T_X$ must be left multiplication by an element of $G$! (Such maps are also compatible with all tensor products, duals, and direct sums.) This result is independent of $K$, so it shows that $$G_{X,B} = \underline{\textrm{Gal}(k/\mathbb{Q})}.$$ Note as well that the calculations we have to carry out are exactly those for the category of representations of $G$ and the universal representation $K[G]$ (where $g$ acts as right multiplication by $g^{-1}$). This situation is also completely dual to the left regular representation; for example, $f$ is dual to the comultiplication map $K[G]\to K[G]\otimes K[G]$ given by $g \mapsto g\otimes g$.
Having said all that, it's time for us to think about the de Rham fiber functor. Automorphisms of this fiber functor are systems $S_X$ of $K$-linear automorphisms $$H_{dR} (X) \otimes K \to H_{dR} (X) \otimes K$$ such that, for all maps of motives $\varphi: Y\to Y'$, $Y, Y' \in \langle X\rangle$, we have a commuting diagram:
\begin{CD}
{H_{dR} (Y) \otimes K} @>{S_Y} >> {H_{dR} (Y) \otimes K}\\
@V H_{dR} (\varphi) VV @V H_{dR} (\varphi) VV\\
H_{dR} (Y') \otimes K @> {S_{Y'}}>> H_{dR} (Y') \otimes K\\
\end{CD}
Recall that the 0th de Rham cohomology is exactly the global sections of the structure sheaf, so for $X = \textrm{Spec } k$ we have $$H_{dR} (X) = \Gamma(\textrm{Spec } k, \mathcal{O}_{\textrm{Spec }k}) = k.$$ Once again, morphisms $X \to X$ in the category of Artin motives are just pullbacks of field automorphisms $g: k \to k$, and now the induced map on $H_{dR}(X) = k$ is just $g$. So our commutative diagram specializes to
\begin{CD}
{k \otimes K} @>{S_X} >> {k \otimes K}\\
@V g \otimes 1 VV @V g\otimes 1 VV\\
{k \otimes K} @>{S_X} >> {k \otimes K}\\
\end{CD}
Now, by the Normal Basis Theorem, there is an element $\alpha \in k$ whose Galois conjugates form a basis for $k$ as a vector space over $\mathbb{Q}$. This means that $k \otimes K$ is isomorphic to $K[G]$ with Galois action by left multiplication -- $cg$ corresponds to $g\alpha \otimes c$ -- and you might think that we were back in the situation for the Betti fiber functor. The difference comes up when we consider the second commutative diagram, that coming from the map $X \times X \to X$ induced by $k \otimes k \to k$. But the key point is that the induced map analogous to the one we called $f$ before, $f': H_{dR}(X) \otimes H_{dR} (X) \to H_{dR} (X)$, is NOT equivalent to $f$ as a map of $G$ representations over $\mathbb{Q}$ (!!!!).
In fact, if $K = \mathbb{Q}$, then there are NO possible $S_X$ except for the identity (unless $G$ is abelian). To see this, note that, by the two commutative diagrams, a valid $S_X$ is a $\mathbb{Q}$-linear map $S: k \to k$ such that $S(xy) = S(x)S(y)$ and $S(gx) = gS(x)$. But if $S$ respects multiplication and addition, it's a field automorphism, so $S(x) = hx$ for some $h \in G$. But then we have $hgx = ghx$ for all $g$, so $h$ is a central element. Only abelian Galois groups can have nontrivial centers, so $T = 1$ on $H_{dR} (X)$ and thus on all motives in $\langle X \rangle$.
On the other hand, if we base change to a large enough $K$, then we do get some $K$-points. Suppose $\alpha$ is a primitive element of $k$ (one whose powers generate $k$ over $\mathbb{Q}$), and call its minimal polynomial $f(\alpha)$. Then $$k \otimes_{\mathbb{Q}} K = \mathbb{Q}[x]/f(x) \otimes_\mathbb{Q} K = K[x]/f(x) = \prod_{\alpha_i} K[x]/(f(x) - \alpha_i) = \prod_{\alpha_i} K.$$ Here, $\alpha_i$ denote the Galois conjugates, and the third isomorphism is by the Chinese Remainder Theorem. The multiplication in this ring (corresponding to $f'$ linearly extended to $K$) looks exactly like the multiplication $f$ on $K[G]$, which we recall is of the form: $$f(c_1g \otimes c_2h) = \begin{cases} 0, & \textrm{if } g\neq h \\ c_1c_2g, & \textrm{if } g = h. \end{cases}$$ Here, $g$ indexes which factor of the product we are in; think of it as the idempotent with a 1 in one index and a 0 in the rest. So after this base change, the arguments in the Betti realization case go through exactly, and we still have $G_{X, dR} (K) = \textrm{Gal} (k/\mathbb{Q})$.
Whew! This is all done more abstractly, more succinctly, and far more canonically in Rosen's paper, but I hope the concrete example is helpful to somebody who stumbles across this question in the future.
