No, there isn't really a loophole. Consider two particles which live in 1D, and let them be bosonic. The wavefunction of the composite system can most generally be written as some symmetric function $\psi(x_1,x_2)=\psi(x_2,x_1)$.
If we make the assumption that
$$\psi(x_1,x_2) = \frac{\psi_A(x_1)\psi_B(x_2) + \psi_B(x_1)\psi_A(x_2)}{\sqrt{2}}$$
and further assume that $\psi_A$ and $\psi_B$ are such that they only have support in regions $R_A$ and $R_B$, $R_A\cap R_B = \emptyset$, then the probability density for the position of the first particle is
$$|\psi_{eff}(x_1)|^2 = \int|\psi(x_1,x_2)|^2 dx_2 = \frac{|\psi_A(x_1)|^2 + |\psi_B(x_1)|^2}{2}$$
and the same holds for $\psi_2$. Therefore, one could imagine a "reduced" wavefunction
$$\phi(x) = \psi_A(x)+\psi_B(x)$$
which has been projected down into 1D space, and whose norm squared gives the probability density for finding a particle (the identity of which is meaningless) at position $x$.
There were two massively-simplifying assumptions here that do not hold in general. First, we assumed that the composite wavefunction could be cleanly separated into single-particle wavefunctions $\psi_A$ and $\psi_B$, which is generally not the case. Consider for example
$$\psi(x_1,x_2) = e^{-(x_1+x_2)^2}$$
For such a wavefunction, the probability of finding the a particle at some $x_1$ is an inextricable function of $x_2$; there is no sense of separate probability distributions for $x_1$ and $x_2$ individually.
Second, we assumed that the single-particle wavefunctions $\psi_A$ and $\psi_B$ were completely separated, in the sense that their respective supports had no overlap. Were this not the case, then the full probability density would not have cleanly separated into positive-definite parts, and the interference effects would become too important to ignore.