I think it's easier to see in Cartesian coordinates. Define the "primitive" moments
\begin{align} q & = \int \rho(\mathbf r)\, \text{d}^3\mathbf{r}\\ p_i & = \int r_i \rho(\mathbf r)\, \text{d}^3 \mathbf{r}\\ Q_{ij} & = \int r_i r_j \rho(\mathbf r)\, \text{d}^3 \mathbf{r} \end{align}
Assume $q = 0$, $p_i$ not all 0.
If you displace the origin by $\mathbf{d}$ and call the new quadrupole moments $Q'_{ij}$, then
\begin{align} Q'_{ij} & = \int (r_i - d_i)(r_j - d_j) \rho \text{d}^3 \mathbf{r}\\ &= Q_{ij} - p_i d_j - p_j d_i + q d_i d_j. \end{align}
If q = 0, finding a translation that makes $Q'_{ij} = 0$ is now a (overspecified) linear problem of six equations in three unknowns. To make it even simpler, assume the dipole is aligned along the z-axis ($p_1 = p_2 = 0$). Then it is easy to see that no translation will change $Q_{11}, Q_{12}, \text{ or } Q_{22}$, so if any of them are nonzero then translating the distribution will not make them zero. If they are all zero, then the quadrupole moment can be entirely zeroed out by setting $d_1 = Q_{13}/p_3$, $d_2 = Q_{23}/p_3$, and $d_3 = Q_{33}/p_3$.
