<B>Tentative.</B>  This is too long for a comment, but it is not yet an answer.  The cokernel of the natural restriction homomorphism is an explicit Galois cohomology group described below.  I believe that this Galois cohomology group vanishes, but I need to review the Galois cohomology of permutation modules and Shapiro's Lemma.  I am writing the derivation below until I can do that.
 
Denote by $S$ the scheme $\text{Spec}\ k$.  Denote by $\mathfrak{g}$ the Galois group $\text{Gal}(\overline{k}/k)$.  Let $\pi:X\to S$ be a smooth, separated, quasi-compact morphism with geometrically irreducible fiber.  Let $\sigma:S\to X$ be a section of $\pi$.  Denote by $G$, resp. $G_\sigma$, the group $S$-scheme that represents $T\mapsto \mathbb{G}_{m,X}(X\times_S T)$, resp. that represents the kernel of $$\sigma^*(T):\mathbb{G}_{m,X}(X\times_S T) \to \mathbb{G}_{m,S}(T).$$  Pullback by $\sigma$ defines a group homomorphism, $$\sigma^*:\text{Br}(X)\to \text{Br}(S).$$  Denote the kernel by $\text{Br}(X)_\sigma$. Pullback by $\pi$ defines a group homomorphism, $$\pi^*:H^2_{\mathfrak{g}}(G_\sigma(\overline{k})) \to \text{Br}(X)_\sigma.$$

<B>Lemma.</B> The cokernel of the natural restriction morphism (1) equals the kernel of $H^2_{\mathfrak{g}}(G_\sigma(\overline{k})) \to \text{Br}(X)_\sigma$.  The group $G_\sigma(\overline{k})$ is a finitely generated, torsion-free Abelian group.  In particular, the cokernel of the natural restriction morphism is a finite group.

<B>Proof.</B> The relevant terms in the Hochschild-Serre spectral sequence give the following long exact sequence, $$0 \to H^1_{\mathfrak{g}}(G(\overline{k})) \to \text{Pic}(X)\xrightarrow{\text{res}} \text{Pic}(X_{\overline{k}})^{\mathfrak{g}} \xrightarrow{\delta} H^2_{\mathfrak{g}}(G(\overline{k})) \to \text{Br}(X).$$  By Hilbert's Theorem 90, the first term is $H^1_{\mathfrak{g}}(G_\sigma(\overline{k}))$.  Pullback by the section $\sigma$ defines a splitting of the map $$H^2_{\mathfrak{g}}(\mathbb{G}_{m,S}(\overline{k}))\to \text{Br}(X).$$  Thus, the cokernel of $\text{res}$ equals the kernel of $\pi^*$.

The group $G_{\sigma}(\overline{k})$ is finitely generated by Roitman (?).  Every torsion-element satisfies an integral equation over $\overline{k}$.  Since $X_{\overline{k}}$ is integral, $\overline{k}$ is integrally closed in $H^0(X_{\overline{k}},\mathcal{O}_X)$.  Thus, $G_{\sigma}(\overline{k})$ is a finitely generated, torsion-free Abelian group.  The profinite Galois cohomology group $H^2_{\mathfrak{g}}(G_\sigma(\overline{k}))$ is torsion.  Thus, $H^2_{\mathfrak{g}}(G_\sigma(\overline{k}))$ is a finite group.  <B>QED</B>