I will prove the stated equality for the case $|\alpha|\leq\pi$. (I believe it is false when $|\alpha|>\pi$.)
The Poisson summation formula tells us that if $f(x)$ is a function and
$$
F(q) = \int_{-\infty}^{+\infty}dx\, f(x)\, e^{-i q x}
$$
is its continuous Fourier transform (FT), then
$$
\sum_{n = -\infty}^{+\infty} f(n)\;=\;
\sum_{n = -\infty}^{+\infty} F(2\pi n)
$$
For this problem, let
$$
f(x) = {\mathrm{sinc}(\alpha x + \theta)}^2\, ,
$$
where $\mathrm{sinc}(x) = \sin(x)/x$. Then the FT of $f(x)$ is
\begin{align*}
F(q)
&= \int_{-\infty}^{+\infty}dx\; {\mathrm{sinc}(\alpha x + \theta)}^2\; e^{-i q x}\\
&= e^{+i \, q\, \theta/\alpha} \int_{-\infty}^{+\infty}dx\; {\mathrm{sinc}(\alpha x)}^2\; e^{-i q x} \qquad \text{Change of variables: $x\rightarrow x - \theta/\alpha$}\\
&= e^{+i \, q\, \theta/\alpha} \frac{1}{|\alpha|}\int_{-\infty}^{+\infty}dz\; {\mathrm{sinc}(z)}^2\; e^{-i u z} \qquad z = |\alpha|x\, , \; u = q/|\alpha|
\end{align*}
The FT of the square of a sinc function is a "tent" function:
\begin{equation*}
\int_{-\infty}^{+\infty}dz\; {\mathrm{sinc}(z)}^2\; e^{-i u z}\;=\;
\begin{cases}
\pi \left(1 - \frac{|u|}{2}\right) & |u| < 2\\
0 & \text{else}
\end{cases}
\end{equation*}
I'm not going to prove this here so as to not clutter up the derivation, but it can be found in various sources, and follows from the convolution theorem for Fourier transforms and the fact that the FT of a sinc function is a "top hat" function.
Putting this all together, we arrive at:
\begin{equation}
F(q) \;=\; e^{+i \, q\, \theta/\alpha} \; \frac{\pi}{|\alpha|} \;
\begin{cases}
1 - \frac{|q|}{2|\alpha|} & |q| < 2|\alpha|\\
0 & \text{else}
\end{cases} \qquad\qquad (1)
\end{equation}
The Poisson summation formula tells us that
\begin{equation*}
\sum_{n = -\infty}^{+\infty} {\left[\frac{\sin(n\alpha + \theta)}{n\alpha + \theta}\right]}^2\;=\;
\sum_{n = -\infty}^{+\infty} f(n)\;=\;
\sum_{n = -\infty}^{+\infty} F(2\pi n)
\end{equation*}
In performing the latter sum over $F(2\pi n)$ using Eq. (1), we see that since $F(q)$ has compact support on $|q| < 2|\alpha|$, we need only sum $n$ over
$|n| < |\alpha|/\pi$. When $|\alpha| \leq \pi$, only the $n = 0$ term contributes to the sum, and we are left with:
\begin{equation*}
\sum_{n = -\infty}^{+\infty} {\left[\frac{\sin(n\alpha + \theta)}{n\alpha + \theta}\right]}^2\;=\;
F(0) \;=\; \frac{\pi}{|\alpha|} \qquad\qquad (|\alpha| \leq \pi)
\end{equation*}
When $|\alpha| > \pi$, more terms with $n \neq 0$ enter the sum, and the expression becomes more complicated, the $\theta$ dependence does not go away, and I'm not sure there's a nice, closed-form expression. For instance, evaluating the sum in Mathematica for $\alpha = 4$ and $\theta = 0$ yields:
\begin{equation*}
\sum_{n = -\infty}^{+\infty} {\left[\frac{\sin(4 n)}{4 n}\right]}^2\;=\;
\frac{6\pi - \pi^2}{8} \;\neq \; \frac{\pi}{4}\, .
\end{equation*}
Edited to add:
The general sum is given by
\begin{equation*}
\sum_{n = -\infty}^{+\infty} {\left[\frac{\sin(n\alpha + \theta)}{n\alpha + \theta}\right]}^2\;=\;
\frac{\pi}{|\alpha|}\left[
1 \;+\; 2\sum_{n = 1}^{\left\lfloor |\alpha|/\pi \right\rfloor} \cos(2\pi \,n\, \theta\, /\, \alpha) \left(1 - \frac{\pi\, n}{|\alpha|}\right)
\right]\, .
\end{equation*}
If this has a "nice" closed form, I'm unaware of it.
Best Answer
You need to be more careful about your antiderivatives: firstly, the cotangent identity shows that $$ \pi \cot{\pi z} - \frac{1}{z} $$ is finite (and hence integrable) in a neighbourhood of the origin (use the comparison test or the integral test or something). Hence, $$ \int_0^z \left(\pi \cot{\pi w} - \frac{1}{w}\right) \, dw \tag{1} $$ exists: an antiderivative is $$ \log{\left( \frac{\sin{\pi w}}{\pi w} \right)}, $$ so the integral (1) is $$ \left[ \log{\left( \frac{\sin{\pi w}}{z} \right)} \right]_0^z = \log{\left( \frac{\sin{\pi z}}{\pi z} \right)} - \log{1} = \log{\left( \frac{\sin{\pi z}}{\pi z} \right)}. $$
Now, for the right-hand side, we observe that it is uniformly convergent on $[0,z]$ for $z$ small enough, so we can integrate term-by-term. We have $$ \int_0^z \frac{2w}{w^2-n^2} \, dw, $$ and as you almost note, $\log{(n^2-w^2)}$ is an antiderivative (note the sign: $z<1$, so $z<n$). However, if we now evaluate the integral, we find $$ \left[ \log{(n^2-w^2)} \right]_0^z = \log{(n^2-z^2)}-\log{(n^2)} = \log{\left( 1-\frac{z^2}{n^2} \right)} $$ Summing these up, exponentiating and multiplying by $\pi z$, we get $$ \sin{\pi z} = \pi z \prod_{n=1}^{\infty} \left( 1-\frac{z^2}{n^2} \right) $$ on $[-1+\varepsilon, 1-\varepsilon]$, for any $\varepsilon>0$, because we can only get uniform convergence here.
The remaining issue is periodicity: you have to check that the right-hand side has period $2$. Extending the identity over the integers is straightforward by taking $\varepsilon \to 0$ and checking continuity there.