For a particle to have physical mass, is it always necessary to have a mass term in the lagrangian?

It is possible for particles to get masses at loop level while they are absent at tree level as long as there is no (non-anomalous) symmetry that forbids it. However, in most models if there is a particle that doesn't have a bare mass then its due to a symmetry which then protects it from getting masses at loop level.

This is often a subtle topic due to chiral symmetry, \begin{equation} \psi \rightarrow e ^{i \gamma_5 \alpha}\psi \end{equation} which can protect fermions from getting masses under loop corrections. This symmetry is broken by a fermionic mass term, $\bar{\psi} \psi $, but can be conserved by the rest of the Lagrangian. In this case if the mass term doesn't appear at tree level (due to some imposed symmetry) it can't appear at higher orders since the chiral symmetry will protect it.

This topic is often discussed in the context of neutrinos.

In addition to JeffDror answer :

In the case of scalar fields, this is definitely the case. In order to get a massless field, you need to fine-tune one parameter of the Lagrangian. In the case of the a $\varphi^4$ theory defined by the parameters ($m_\Lambda$, $g_\Lambda$, $\Lambda$), i.e. the bare mass, interaction and the UV cut-off, one need to fine-tune one of the parameter (usually one chooses $m_\Lambda$) to insure that the renormalized mass $m_R=0$.

Note that generically the correction to the mass is positive, and one needs $m_\Lambda^2<0$.