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First take the trace in four spacetime dimensions:The product rule gives $\square {\varphi }^2=2(\partial \varphi {)}^2+2\varphi \square \varphi$. For the massless equation of motion, $\varphi \square \varphi =-\lambda {\varphi }^4/6$, soBoth coefficients vanish forThe minus sign follows from the order ${\partial }^{\mu }{\partial }^{\nu }-{\eta }^{\mu \nu }\square$ used in the question's stress-energy tensor improvement. Commuting partial derivatives shows that this improvement has identically zero divergence, so it preserves the conservation law.

The precise trace identity, without using the equation of motion, isHence the massless tensor is traceless on shell. It is not traceless for arbitrary off shell configurations: a nonzero constant $\varphi =A$ with $\lambda \ne 0$ has ${\Omega }^{\mu }{}_{\mu }=\lambda A^4/6$. This is a classical result; it does not assert absence of a quantum trace anomaly.

A trace anomaly is a quantum contribution to the trace of a stress-energy tensor which was traceless on shell in the classical theory. Running renormalized couplings can supply such a contribution through their beta functions; curvature-dependent contributions can occur as well. Classical stress-energy tensor improvement alone does not remove quantum scale dependence.