= Solution
Use units $c=1$ and signature $(+---)$. For a massive <test particle>, write its <four-momentum> as $p^i=mu^i$, where $u^i=dx^i/d\tau$, $\tau$ is Minkowski <proper time>, and $u^iu_i=1$. The stated <flat-background scalar-force theory of gravity> then gives
$$
\frac{dp^i}{d\tau}=m\left(\partial^i\Phi-u^i u^j\partial_j\Phi\right).
$$
Contract with $p_i=mu_i$. The two terms cancel:
$$
\frac{d(m^2)}{d\tau}=2p_i\frac{dp^i}{d\tau}
=2m^2\left(u^j\partial_j\Phi-u^iu_i u^j\partial_j\Phi\right)=0.
$$
Thus <rest mass> is conserved and the <acceleration> law is
$$
\boxed{\frac{du^i}{d\tau}=\partial^i\Phi-u^i(u\cdot\partial\Phi).}
$$
There is no mass or composition parameter in this equation. Particles released at the same event with the same <velocity> consequently have identical trajectories. In a static weak field at small speed, its spatial part becomes $d^2\mathbf x/dt^2=-\boldsymbol\nabla\Phi$, since raising a spatial index introduces a minus sign. The field equation likewise reduces to $\nabla^2\Phi=4\pi G\rho$ for nonrelativistic matter, because $T^i{}_i\simeq\rho$ and $\Box=-\nabla^2$ in the static limit.
Therefore \b[yes: the model agrees with the Eötvös test of universal free fall]. In the point-particle/test-body approximation, inertial and passive gravitational masses have the same universal ratio. This is the <weak equivalence principle> tested by the <Eötvös experiment>; it is not a claim about strongly self-gravitating bodies or a proof of all aspects of the <Einstein equivalence principle>.
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