No.
The natural objects in Minkowski space are Lorentz scalars, 4-vectors, etc, and there is no room for relativistic mass there.
The definition of relativistic mass is "a different mass ($\gamma m$) so that Newtonian formula (e.g $p=mv$) work in regions where Newtonian formula don't work". Definitely sus.
The relevant Minkowski space things are:
$$ u_{\mu} = (\gamma c, \gamma \vec v) $$
(Note: $\gamma$ is here, and mass is not). Then
$$ p_{\mu} = mu_{\mu}$$
You can also write:
(note: looks like a generalization of $\vec p = m\vec v$...is that not good enough?).
$$p^2 = m^2|u|^2 = (mc)^2 $$
We're done: no relativistic mass.
You can also write:
$$ p_{\mu} = mu_{\mu} = (\gamma mc, \gamma m\vec v)$$
and then call that:
$$ p_{\mu} = (E/c, \vec p) $$
with
$$ E = \gamma mc^2 $$ $$ \vec p = \gamma m \vec v$$
and then have the urge to do:
$$ m \rightarrow \gamma m $$
so:
$$ E = mc^2 $$ $$ \vec p = m\vec v $$
at all $v$, but that has nothing to do with Minkowski space.