For a $3\times 3$ symmetric matrix $Q$, one can construct a map to $SO(3)$ which sends $Q$ to a matrix which diagonalizes it. If $Q$ has distinct eigenvalues, there are three choices for rotation matrix: given one (guaranteed by the spectral theorem) you can always cyclically permute the columns to get two more. Choosing one arbitrarily defines a locally smooth map. That is, perturbing the entries of $Q$ slightly will only slightly perturb the eigenvalues and eigenvectors.
However, if $Q$ has repeated eigenvalues then given $R \in SO(3)$ which diagonalizes $Q$, we also have that $R'(\theta) R$ is a valid choice, where $R'(\theta)$ is a rotation by any angle $\theta$ in the plane spanned by the eigenvectors corresponding to repeated eigenvalues. It seems reasonable to me that if we arbitrarily choose some $R$, then slightly perturbing $Q$ could give another element of $SO(3)$ which is far away from $R$ but perhaps close to $R'(\theta') R$ for some $\theta'$.
The question is then, is it possible to construct a mapping so that no such discontinuities occur, or will such a mapping always be discontinuous at elements with repeated eigenvalues?
Yes, discontinuities at repeated eigenvalues are essential for such a map. If $Q_0$ has a repeated eigenvalue and thus freedom to choose $R'(\theta)R,$ then for any choice of $\theta$ you can find $Q$ arbitrarily close to $Q_0$ such that $R'(\theta)R$ is the only valid choice to diagonalize $Q.$
For a concrete example, if the two-dimensional eigenspace is spanned by $e_1$ and $e_2$ (and the third eigenvalue is distinct) then $$Q_1 = Q_0 + \left(\begin{matrix} \epsilon & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\right), Q_2 = Q_0 + \left(\begin{matrix} \epsilon & -\epsilon & 0 \\ -\epsilon & \epsilon & 0 \\ 0 & 0 & 0 \end{matrix}\right) $$
are both $4\epsilon$-close to $Q_0$ but have unique diagonalizing bases that are a full $\pi/4$ turn apart.