The details of what I will say can be found in standard texts on the subject(See for example Peskin and Schroeder:Pg 284), but I will emphasize the main ideas, since some of the ideas may be obscured in calculations sometimes.

You are computing the transition amplitude:
$$Z[J]=\lim_{t,t' \to \mp \infty}\frac{\langle q't'|q,t \rangle_J}{\langle q't'|q,t \rangle}$$
Where is it easy to see that this expression is written out in the Heisenberg picture(Operator $Q_{H}(t)$ has time dependance which its Eigenstates($|q,t \rangle,|q',t' \rangle$ inherit).

Written in the Schrödinger picture: 
$$Z[J]=\lim_{t,t' \to \mp \infty}\frac{\langle q'|e^{-i\int_{t}^{t'}dt''(H-QJ(t''))}|q \rangle}{\langle q'|e^{-iH(t-t')}|q \rangle}$$
From here on out, one uses the path integral ideas to convert this to a path integral of the form:
$$Z[J]=\lim_{t,t' \to \mp \infty}\frac{\int D[q] e^{i\int_{t}^{t'}dt''\frac{m}{2}\dot{q}^2-V(q)+qJ(t'')}}{\int D[q] e^{i\int_{t}^{t'}dt''\frac{m}{2}\dot{q}^2-V(q)}}$$
Its easy to see that both numerator and denominator are oscillating integrals whose convergence is not guaranteed. So the trick is to make the prescription: $m \to m+ i\epsilon, V \to V -i\epsilon V(q)$. This creates a damping factor for the path integral of the form: $exp^{-\epsilon \int_t^{t'}dt'' H(p,q)}$. Provided that the hamiltonian is bounded from below(existence of a stable vacuum),this is a positive damping factor. What is the effect of this prescription on the original partition function?
To see this, consider the original matrix element, now written out explicitly:
$$\langle q'|e^{-i\left(\frac{p^2}{2m+2i\epsilon}+V(q)(1-i\epsilon)\right)(t'-t)}|q \rangle$$
$$=\langle q'|e^{-i\left(\frac{p^2}{2m}(1-\frac{i\epsilon}{m})+V(q)(1-i\epsilon)\right)(t'-t)}|q \rangle=\langle q'|e^{-i\left(\frac{p^2}{2m}+V(q)\right)(1-i\epsilon)(t'-t)}|q \rangle$$
Now expand the position eigenstates in the basis of hamiltonian Eigenstates in a standard fashion:
$$|q\rangle=\sum_i c_i(q) |n_i\rangle$$ 
Where $|n_i\rangle$ anre the energy eigenstates.
Some algebra later, the matrix element turns out to be:
$$=\langle q'|e^{-i\left(\frac{p^2}{2m+2i\epsilon}+V(q)(1-i\epsilon)\right)(t'-t)}|q \rangle =\sum_i e^{E_i(1-i\epsilon)(t'-t)}c_i(q)c_i^*(q')$$
Pulling out an $E_0$:
$$=e^{E_0(1-i\epsilon)(t'-t)}\left(c_0(q)c_0^*(q')+\sum_{i\neq 0} e^{(E_i-E_0)(1-i\epsilon)(t'-t)}c_i(q)c_i^*(q')\right) $$
Similar algebra goes through for the numerator with the $J$'s.Now, in the $t',t\to \pm \infty$ all but the first term goes to zero. Further, the factors of $e^{E_0(1-i\epsilon)(t'-t)}\left(c_0(q)c_0^*(q')\right)$ cancel in the numerator and denominator and therefore only the vacuum to vacuum transition amplitude the presence of a source and the vacuum to vacuum transition amplitude in the absence of a sources remain in the numerator and denominator respectively. Notice that all the information about the boundary conditions are in the $c_0(q)$,$c_0(q')$. But these cancel meaning that the boundary conditions applied don't really matter.

Hope this helps!