We introduce $$f(z) = \pi \csc(\pi z) \frac{P(z)}{Q(z)} $$ and in order to evaluate the sum in question using residues as proposed, we need to show that the integral along the square goes to zero. To do this, assume the degree of $Q(z)$ is some integer $m$ plus the degree of $P(z),$ where $m$ is at least two.
I will do two of the four sides of the contour. On the left side, call it $\Gamma_1$, we have $z = N + \frac{1}{2} + it$, with $-(N+\frac{1}{2}) \le t \le N+\frac{1}{2}.$
This yields
$$\left|\int_{\Gamma_1} f(z) dz \right| \le (2N+1) \max_{\Gamma_1} |f(z)|.$$
Now $$|\csc(\pi z)| =
\frac{2}
{\left|e^{\pi i (N + \frac{1}{2}) -\pi t}-e^{- \pi i (N + \frac{1}{2}) +\pi t}\right|}
= \frac{2}{\left|(-1)^N i e^{-\pi t} + (-1)^N i e^{\pi t}\right|} =
\frac{2}{\left|e^{-\pi t} + e^{\pi t}\right|}\le 1,$$
because the maximum of the norm occurs at $t=0,$ where the denominator is minimal.
It follows that the bound on the integral is in fact
$$(2N+1) \max_{\Gamma_1} |f(z)| \le
(2N+1) \max_{\Gamma_1} \left|\frac{P(z)}{Q(z)}\right| \in
\Theta\left((2N+1)\frac{1}{N^m}\right).$$
But this is$$\Theta\left(\frac{1}{N^{m-1}}\right)
\to 0 \quad \text{as} \quad N\to\infty.$$
On the top side, call it $\Gamma_2$, we have $z = t + i(N + \frac{1}{2})$, with $-(N+\frac{1}{2}) \le t \le N+\frac{1}{2}.$
This yields
$$\left|\int_{\Gamma_2} f(z) dz \right| \le (2N+1) \max_{\Gamma_2} |f(z)|.$$
The cosecant term along $\Gamma_2$ has
$$|\csc(\pi z)| =
\frac{2}
{\left|e^{\pi i t -\pi (N + \frac{1}{2})}-e^{- \pi i t +\pi (N + \frac{1}{2})}\right|}
\le \frac{2}{e^{\pi (N + \frac{1}{2})} - e^{-\pi (N + \frac{1}{2})}}
\in \Theta\left(e^{-\pi N}\right).$$
It follows that the bound on the second integral is
$$(2N+1) \max_{\Gamma_2} |f(z)| \le
(2N+1) \frac{2}{e^{\pi (N + \frac{1}{2})} - e^{-\pi (N + \frac{1}{2})}}
\max_{\Gamma_2} \left|\frac{P(z)}{Q(z)}\right| \in
\Theta\left((2N+1) e^{-\pi N}\frac{1}{N^m}\right).$$
But this is$$\Theta\left(\frac{e^{-\pi N}}{N^{m-1}}\right)
\to 0 \quad \text{as} \quad N\to\infty.$$
This concludes the argument and shows that
$$ \sum_{n=-\infty}^\infty (-1)^n \frac{P(n)}{Q(n)} =
-\sum_{z_0\in S}
\operatorname{Res}\left(\pi\csc(\pi z) \frac{P(z)}{Q(z)}; z=z_0\right),$$
where $S$ is the set of poles of $f(z)$ other than at the integers and we assume for simplicity that the quotient of the two polynomials does not have any poles at the integers.
Here we have made use of the fact that with
$$P(n) = p_0 + p_1 n + p_2 n^2 + \cdots + p_\alpha n^\alpha$$ and
$$Q(n) = q_0 + q_1 n + q_2 n^2 + \cdots + q_\beta n^\beta$$
and $\beta\ge\alpha,$ we have
$$ \frac{P(n)}{Q(n)} =
\frac{p_\alpha + \cdots + p_0/n^\alpha}{q_\beta n^{\beta-\alpha} + \cdots + q_0/n^\alpha}
\in \Theta\left( \frac{1}{n^{\beta-\alpha}} \right)
\quad \text{as} \quad n \to\infty.$$
