Let us take $k=\mathbf{Q}$ and $M=\mu_3$. Then $H^0(\Gamma,M)=0$. On the other hand, $M^D\simeq \mathbf{Z}/3\mathbf{Z}$, and one can show that $H^2(\Gamma,M^D)\neq 0$. It follows that there exists an element $x\neq 0$ in $H^2(\Gamma,M^D)$, and of course we have $$x^0=0\in {\rm Hom}( H^0(\Gamma,M), {\rm Br}(k))$$ because $H^0(\Gamma,M)=0$. (We thank Ofer Gabber for this counter-example.)

Alternatively, there exists a field $k$ of characteristic 0 such that ${\rm Br}(k)=0$, but $k$ is not of dimension $\le 1$; see Serre, Galois Cohomology, II.3.1, Exercise 1. For such $k$, there exists a finite $\Gamma$-module $A$ such that $H^2(\Gamma,A)\neq 0$. Take $M=A^D$; then $A=M^D$. It follows that there exists an element $x\neq 0$ in $H^2(\Gamma,M^D)$, and of course we have $$x^0=0\in {\rm Hom}( H^0(\Gamma,M), {\rm Br}(k))$$ because ${\rm Br}(k)=0$. (We thank David Harari for this counter-example.)

Let us take $k=\mathbf{Q}$ and $M=\mu_3$. Then $H^0(\Gamma,M)=0$. On the other hand, $M^D\simeq \mathbf{Z}/3\mathbf{Z}$, and one can show that $H^2(\Gamma,M^D)\neq 0$. It follows that there exists an element $x\neq 0$ in $H^2(\Gamma,M^D)$, and of course we have $$x^0=0\in {\rm Hom}( H^0(\Gamma,M), {\rm Br}(k))$$ because $H^0(\Gamma,M)=0$. (We thank Ofer Gabber for this example.)

Alternatively, there exists a field $k$ of characteristic 0 such that ${\rm Br}(k)=0$, but $k$ is not of dimension $\le 1$; see Serre, Galois Cohomology, II.3.1, Exercise 1. For such $k$, there exists a finite $\Gamma$-module $A$ such that $H^2(\Gamma,A)\neq 0$. Take $M=A^D$; then $A=M^D$. It follows that there exists an element $x\neq 0$ in $H^2(\Gamma,M^D)$, and of course we have $$x^0=0\in {\rm Hom}( H^0(\Gamma,M), {\rm Br}(k))$$ because ${\rm Br}(k)=0$. (We thank David Harari for this example.)

(Thanks to an idea of Ofer Gabber.) Take $k={\bf Q}$, $M=\mu_3$; then clearly $H^0({\bf Q},M)=0$. We have $M^D\simeq{\bf Z}/3{\bf Z}$. It suffices to show that $H^2({\bf Q},{\bf Z}/3{\bf Z})\neq 0$. Indeed, then we can take any nonzero element $x\in H^2({\bf Q},M^D)$. Since $H^0({\bf Q},M)=0$, we will have $x^0=0$, as required.

We show that $H^2({\bf Q},{\bf Z}/3{\bf Z})\neq 0$. It is not hard to show that $$ H^2({\bf Q},{\bf Z}/3{\bf Z})\cong\bigoplus H^2({\bf Q}_v, {\bf Z}/3{\bf Z}).$$ Therefore, it suffices to show that $H^2({\bf Q}_p,{\bf Z}/3{\bf Z})\neq 0$ for some prime $p$.

We take $p=7$. We have $$\mu_3=\langle\zeta\rangle,\quad\text{where } \zeta=\tfrac{-1+\sqrt{-3}}{2}.$$ We have $$-3\equiv 2^2\pmod{7},$$ and therefore, $-3$ is a square in ${\bf Q}_7$. It follows that ${\bf Q}_7(\zeta)={\bf Q}_7$, and hence, ${\bf Z}/3{\bf Z}\simeq \mu_3$ over ${\bf Q}_7$. It remains to notice that $$H^2({\bf Q}_7,\mu_3)\cong {\rm Br}({\bf Q}_7)_3=\tfrac{1}{3}{\bf Z}/{\bf Z}\neq 0.$$

Let us take $k=\mathbf{Q}$ and $M=\mu_3$. Then $H^0(\Gamma,M)=0$. On the other hand, $M^D\simeq \mathbf{Z}/3\mathbf{Z}$, and one can show that $H^2(\Gamma,M^D)\neq 0$. It follows that there exists an element $x\neq 0$ in $H^2(\Gamma,M^D)$, and of course we have $$x^0=0\in {\rm Hom}( H^0(\Gamma,M), {\rm Br}(k))$$ because $H^0(\Gamma,M)=0$. (We thank Ofer Gabber for this counter-example.)

Alternatively, there exists a field $k$ of characteristic 0 such that ${\rm Br}(k)=0$, but $k$ is not of dimension $\le 1$; see Serre, Galois Cohomology, II.3.1, Exercise 1. For such $k$, there exists a finite $\Gamma$-module $A$ such that $H^2(\Gamma,A)\neq 0$. Take $M=A^D$; then $A=M^D$. It follows that there exists an element $x\neq 0$ in $H^2(\Gamma,M^D)$, and of course we have $$x^0=0\in {\rm Hom}( H^0(\Gamma,M), {\rm Br}(k))$$ because ${\rm Br}(k)=0$. (We thank David Harari for this counter-example.)