$3D$ rotation matrix uniqueness

Let's assume the matrix $R$ for a given rotation is not unique and there exists another matrix $Q\neq R$ which performs the exact same rotation. This means

$\forall \mathbf{v}\in\mathbb{R}^n:\quad Q\mathbf{v}=R\mathbf{v}$

and thus

$\forall \mathbf{v}\in\mathbb{R}^n:\quad (Q-R)\mathbf{v}=\mathbf{0}$

This means, that the null-space of $M=(Q-R)$ (the sub-space of all vectors whose image is the zero-vector) has to consist of the whole $\mathbb{R}^n$ (and therefore has a dimension of $n$, of course). The rank-nullity-theorem now states, that the rank of $M$ is the difference of its dimension and the dimension of its null-space, in this case $0$. But then again, only the zero matrix has a rank of $0$.

This means the above equation only holds for $(Q-R)=O$ (with $O$ being the $n\times n$ zero matrix), which in turn implies $Q=R$. This contradicts the assumption. So each rotation (in fact any linear transformation) in $\mathbb{R}^n$ corresponds to a unique $n\times n$ matrix (for a given base $B$, of course). Moreover each orthogonal matrix $R\in\mathbb{R}^{n\times n}$ with $\det R=1$ represents a unique rotation in $\mathbb{R}^n$ (again for a given base $B$ of $\mathbb{R}^n$).

In fact the matrix representation is even more unique than the axis-angle or quaternion representation (not to speak of the ambiguities of Euler angles), since because of the periodicity of the $\sin$ and $\cos$ used in the matrix representation you don't have the angle ambiguity, because $R(\theta)=R(\theta+2\pi k)$. And you don't even have the ambuigity of negated axis and angle (or negated quaternion) being the same rotation (which you especially outruled in your question), because their matrix represntations are in fact the same.