$i\varepsilon$ in momentum space propagator; is it actually needed? In (say) phi-4 theory the momentum space propagator is given by:
$$\frac{i}{p^2-m^2+i\varepsilon}$$
where I am using the signature $(+---)$. Now momentum space we can do momentum space integrals using the Schwinger Paramterization etc in which we do not take any contour integrals etc and as far as I could tell 
$$\frac{i}{p^2-m^2-i\varepsilon}$$
would give exactly the same results. 
My question is therefore when is it valid to take $\varepsilon \rightarrow 0$ in momentum space and why would we keep it?
 A: Yes, it is necessary. See, the real space propagator isn't actually analytic, and the poles in the momentum space propagator will make of your integrals diverge unless you move them off of the integration axis (equivalently, deform the contour around the poles) and take a limit that moves them back onto the integration axis. Because the real space propagator isn't completely analytic, the way you move the poles off of the integration axis in momentum space will affect which type of propagator you're using.
To see how this happens, take the Fourier transform version of the real space propagator
\begin{align}G(x,y) &= \int \frac{\operatorname{d}^4 p \operatorname{d}^4p'}{(2\pi)^4} \left(\frac{\delta^4(p-p')}{p^\mu p_\mu-m^2}\right) \mathrm{e}^{-ip^\mu (x_\mu-y_\mu)}\\
   &= \int \frac{\operatorname{d}^3 p }{(2\pi)^4} \int_{-\infty}^\infty \operatorname{d}p^0 \left(\frac{1}{p^0p^0 - p^ip^i-m^2}\right) \mathrm{e}^{-ip^\mu (x_\mu-y_\mu)}.
\end{align}
Notice how the $p^0$ integration has two order $1$ poles in it at $p^0 = \pm \sqrt{p^ip^i + m^2}$. This makes the $p^0$ integral, as written, ill defined because it basically involves subtracting two infinite quantities. Thus, the integral only becomes meaningful if we have some prescription for how the limits approach the poles (e.g. Cauchy principle value) or we move the poles off of the integration axis. 
With the Feynman $i\epsilon$ prescription, one pole moves up and the other down, leading the residue theorem to give
$$G(x,y) = \ldots + \int \frac{\operatorname{d}^3 p }{(2\pi)^4} \left(\frac{2\cos\left(\left[x^0-y^0\right]\sqrt{\mathbf{p}\cdot\mathbf{p} + m^2}\right)}{\sqrt{ \mathbf{p}\cdot\mathbf{p} + m^2}}\right) \mathrm{e}^{i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})}.$$
If you use an $i\epsilon$ prescription that moves both poles in the same direction, e.g. $p^0 \rightarrow p^0+i\epsilon$, then you get something like
$$G(x,y) = \ldots + \int \frac{\operatorname{d}^3 p }{(2\pi)^4} \left(\frac{2i\sin\left(\left[x^0-y^0\right]\sqrt{\mathbf{p}\cdot\mathbf{p} + m^2}\right)}{\sqrt{ \mathbf{p}\cdot\mathbf{p} + m^2}}\right) \mathrm{e}^{i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})} \Theta\left(x^0-y^0\right).$$
Note that I've not been very careful, so there may be some sign errors or incorrect leading numerical factors. I've also hidden the expected light cone delta functions inside of the "$\ldots$" because I do not yet have the background in distribution theory needed to explain how they come from the integral in question.
Bottom line - while the $i\epsilon$ prescription doesn't make a difference to momentum space integrals, it defines which version of the propagator you're using in real space so it should be kept around until that transition in case you, or subsequent users of your work, want to make that transition at some point.
See also the Wikipedia propagator article.
