Yes, it is. This follows from the theory of central simple algebras. Here's another proof, possibly similar to Aakumadula's: let $D'=D\otimes \bar{k}\cong GL_2(\bar{k})$. If $D'$ has basis $1,a$ then it's commutative; otherwise, it has basis $1,a,b$. I claim that $a,b$ have a common eigenbasis in $\bar{k}^2$. Indeed, suppose not. Then $a,b$ are diagonalizable. Pick a basis in which $a$ is diagonal. Then $b$ won't be either upper- or lower-triangular, and you can see combinatorially that $1,a,b,ab$ are linearly independent, so $D'=Q'$ and $D=Q$. Thus $D'$ is the space of upper-triangular matrices in some basis. But this is impossible, e.g. since then $D'$ has an ideal fixed by the Galois action.

Yes, it is. This follows from the theory of central simple algebras. Here's another proof, possibly similar to Aakumadula's: let $D'=D\otimes \bar{k}\cong GL_2(\bar{k})$. If $D'$ has basis $1,a$ then it's commutative; otherwise, it has basis $1,a,b$. I claim that $a,b$ have a common eigenvalue in $\bar{k}^2$. Indeed, suppose not. Then $a,b$ are diagonalizable. Pick a basis in which $a$ is diagonal. Then $b$ won't be either upper- or lower-triangular, and you can see combinatorially that $1,a,b,ab$ are linearly independent, so $D'=Q'$ and $D=Q$. Thus $D'$ is the space of upper-triangular matrices in some basis. But this is impossible, e.g. since then $D'$ has an ideal fixed by the Galois action.