This answer continues the line of arguments of my Phys.SE answer here, where it was argued that the Lagrangian $L(\vec{v})$ is a function of velocity $\vec{v}$ only.

To implement isotropy in space as a quasi-symmetry of the Lagrangian consider an infinitesimal rotation $$v^i\quad \longrightarrow\quad v^{\prime t}~=~v^i+ \epsilon^{ij}v^j, \tag{1}$$ where $\epsilon^{ij}=-\epsilon^{ji}$ is an infinitesimal antisymmetric matrix.

So the infinitesimal change of the Lagrangian $$\Delta L~:=~L^\prime-L ~=~ \frac{\partial L}{\partial v^i}\epsilon^{ij}v^j\tag{2}$$ should be a total time derivative $$ \frac{\mathrm{d}F}{\mathrm{d}t} \tag{3}$$ even off-shell. Since eq. (2) only depends on the velocity $\vec{v}$, it follows that $$F~=~\vec{a}\cdot\vec{q}\tag{4}$$ is a linear function of position. We now decompose the Lagrangian $$L(\vec{v})~=~L_1(\vec{v})+L_{\neq 1}(\vec{v}) \tag{5}$$ in a linear and a non-linear part. The linear part $L_1(\vec{v})$ is a total time derivative, so we can w.l.o.g. assume that $L_1(\vec{v})=0$ is zero. Then the function $F=0$ also becomes zero.

So we can w.l.o.g. assume that rotations are implemented as strict symmetries (rather than just quasi-symmetries) of the Lagrangian, i.e. the Lagrangian $L=L(v^2)$ depends only on the speed.

This answer continues the line of arguments of my Phys.SE answer here, where it was argued that the Lagrangian $L(\vec{v})$ is a function of velocity $\vec{v}$ only.

To implement isotropy in space as a quasi-symmetry of the Lagrangian consider an infinitesimal rotation $$v^i\quad \longrightarrow\quad v^{\prime i}~=~v^i+ \epsilon^{ij}v^j, \tag{1}$$ where $\epsilon^{ij}=-\epsilon^{ji}$ is an infinitesimal antisymmetric matrix. (Technically it belongs to the Lie algebra $so(3)$ of the 3D rotation group $SO(3)$.)

So the infinitesimal change of the Lagrangian $$\Delta L~:=~L^\prime-L ~=~ \frac{\partial L}{\partial v^i}\epsilon^{ij}v^j\tag{2}$$ should be a total time derivative $$ \frac{\mathrm{d}F}{\mathrm{d}t} \tag{3}$$ even off-shell. Since eq. (2) only depends on the velocity $\vec{v}$, it follows that $$F~=~\vec{a}\cdot\vec{q}\tag{4}$$ is a linear function of position $\vec{q}$. We now decompose the Lagrangian $$L(\vec{v})~=~L_1(\vec{v})+L_{\neq 1}(\vec{v}) \tag{5}$$ into a linear and a non-linear part. OP's example belongs to the linear part $L_1(\vec{v})$. The linear part $L_1(\vec{v})$ is a total time derivative, so we can w.l.o.g. assume that $L_1(\vec{v})=0$ is zero. Then the function $F=0$ also becomes zero.

So we can w.l.o.g. assume that rotations are implemented as strict symmetries (rather than just quasi-symmetries) of the Lagrangian, i.e. the Lagrangian $L=L(v^2)$ depends only on the speed.