Martin is right. It is very simple to see it: just consider for a moment to add this extra superpotential term $$ \delta W(\Phi_i)=\frac{1}{2}m_{i j}\Phi_i \Phi_j $$ where $\Phi_i$ are chiral superfields. This extra $W$ corresponds to an extra lagrangian term $$ \delta\mathcal{L}=-\frac{1}{2}m_{i j}(\psi_i\psi_j+\text{h.c.})- \bar{\phi}_iM^2_{ij}\phi_j\qquad M^2_{ij}=\bar{m}_{ik}m_{jk}\,. $$ Therefore, starting with $W$ and adding a soft mass for the fermions $\delta\mathcal{L}_{\mathrm{soft}-\psi}=-\frac{1}{2}m_{i j}(\psi_i\psi_j+\text{h.c.})$ $$ W(\Phi)\rightarrow \mathcal{L}=\mathcal{L}_{SUSY}+\delta\mathcal{L}_{\mathrm{soft}-\psi} $$ is just the same as doing starting with $W+\delta W$ (with $\delta W$ in the first equation above) and add a soft mass terms for the scalars, $\delta\mathcal{L}_{\mathrm{soft}-\phi}=+\bar{\phi}_iM^2_{ij}\phi_j$ $$ W(\Phi)+\delta W(\Phi)\rightarrow \mathcal{L}=\mathcal{L}^\prime_{SUSY}+\delta\mathcal{L}_{\mathrm{soft}-\phi}=\mathcal{L}_{SUSY}+\delta\mathcal{L}_{\mathrm{soft}-\psi} $$
In other words, it is intuitive that only the mass splitting between fermions and boson that matters and should be considered. Adding soft masses that respect susy, that is that do not split fermions and bosons, it is just like adding a mass term to $W$. Therefore, the mass term for the fermion is soft, in the sense that does not give rise to quadratic divergences, since it is equivalent to a new superpotential with soft scalar masses that we know do not generate quadratic divergences. Perhaps, the only possible subtle point left that I see would be about the vacuum stability, since the soft masses for the scalars come flip in sign. But in any case, this is not about their being soft.
