I've found an integral representation for my series: With $$S(z,a;q):=\frac{\pi q\varphi(a,q)}{\varphi'(0,q)}\sum_{k\in\mathbb Z}\frac{e^{-2(a+k)\pi qz}}{\sinh(a+k)\pi q},$$
\begin{align}\int_0^{i/q}S(z,a;q)\,dz&=\frac{\pi q\varphi(a,q)}{\varphi'(0,q)}\int_0^{i/q}\sum_{k\in\mathbb Z}\frac{e^{-2(a+k)\pi qz}}{\sinh(a+k)\pi q}\,dz\\[2ex]
&=\frac{\pi q\varphi(a,q)}{\varphi'(0,q)}\sum_{k\in\mathbb Z}\frac1{\sinh(a+k)\pi q}\frac{e^{-2(a+k)i\pi}-1}{-2(a+k)\pi q}\\[2ex]
&=\frac{\varphi(a,q)}{2\varphi'(0,q)}\sum_{k\in\mathbb Z}\frac{e^{-2ai\pi-2ki\pi}-1}{-(a+k)\sinh(a+k)\pi q}\\[2ex]
&=\frac{\varphi(a,q)}{2\varphi'(0,q)}(1-e^{-2ai\pi})\sum_{k\in\mathbb Z}\frac1{(a+k)\sinh(a+k)\pi q}\tag1.
\end{align}And $S(z,a;q)$ just happens to be the quotient of two theta functions (see proof in the appendix below): $S=\phi(z+a,q)/\phi(z,q)$, where
\begin{align}\phi(z,q):&=\sum_{k\in\mathbb Z}(-1)^ke^{-\pi q(k+z)^2}\\[2ex]
&=e^{-\pi qz^2}\prod_{k\ge1}(1-e^{-2k\pi q})(1-e^{-(2k-1)\pi q+2\pi qz})(1-e^{-(2k-1)\pi q-2\pi qz})\tag2\\[3ex]
\phi(z,q)&=\varphi(z\!+\!1/2,q)
\end{align}
The identity $\phi(z+i/q,q)=e^{\pi/q-2i\pi z}\phi(z,q)$ (which can be derived directly from $(2)$) leads to $S(z+i/q,a;q)=e^{-2ai\pi}S(z,a;q)$, implying double periodicity for $a\in\mathbb Q$. Thus the expression $(1)$, in principle, will have a closed form in terms of log-theta functions when $a$ is rational.
Proof that $S(z,a;q)$ is a quotient of theta functions
To prove $S$ has such a representation, create a quotient product with $(2)$:\begin{align}\frac{\phi(z+a,q)}{\phi(z,q)}&=e^{-\pi q(z+a)^2+\pi qz^2}\prod_{k\ge1}\bigg(\frac{1-e^{-(2k-1)\pi q+2\pi q(z+a)}}{1-e^{-(2k-1)\pi q+2\pi qz}}\bigg)\bigg(\frac{1-e^{-(2k-1)\pi q-2\pi q(z+a)}}{1-e^{-(2k-1)\pi q-2\pi qz}}\bigg)\\[3ex]
&=e^{-2a\pi qz-a^2\pi q}\prod_{k\ge1}\bigg(\frac{1-e^{-(2k-1)\pi q+2\pi qz}\cdot e^{2a\pi q}}{1-e^{-(2k-1)\pi q+2\pi qz}}\bigg)\bigg(\frac{1-e^{-(2k-1)\pi q-2\pi qz}\cdot e^{-2a\pi q}}{1-e^{-(2k-1)\pi q-2\pi qz}}\bigg)\\[3ex]
&=e^{-2a\pi qz-a^2\pi q}\prod_{k\ge1}\bigg(e^{2a\pi q}+\frac{1-e^{2a\pi q}}{1-e^{-(2k-1)\pi q+2\pi qz}}\bigg)\bigg(e^{-2a\pi q}+\frac{1-e^{-2a\pi q}}{1-e^{-(2k-1)\pi q-2\pi qz}}\bigg)
\end{align}Expanding the product will produce a constant (w.r.t. $z$) and terms with denominators $(1-e^{-(2k-1)\pi q+2\pi qz})(1-e^{-(2n-1)\pi q-2\pi qz})$, and using partial fraction decomposition,\begin{align}\frac1{(1-e^{-(2k-1)\pi q+2\pi qz})(1-e^{-(2n-1)\pi q-2\pi qz})}&=\frac{e^{(2n-1)\pi q+2\pi qz}}{(1-e^{-(2k-1)\pi q+2\pi qz})(e^{(2n-1)\pi q+2\pi qz}-1)}\\[4ex]
&=\frac1{1-e^{-(2k-1)\pi q+2\pi qz}}+\frac1{(1-e^{-(2k-1)\pi q+2\pi qz})(e^{(2n-1)\pi q+2\pi qz}-1)}\\[3ex]
&=\frac1{1-e^{-(2k-1)\pi q+2\pi qz}}\\
&\quad+\frac1{e^{-(2n-1)\pi q}-e^{(2k-1)\pi q}}\bigg(\frac{e^{-(2n-1)\pi q}}{1-e^{-(2k-1)\pi q+2\pi qz}}-\frac{e^{(2k-1)\pi q}}{1-e^{(2n-1)\pi q+2\pi qz}}\bigg)\\[2ex]
&\cong A+\frac B{1-e^{-(2k-1)\pi q+2\pi qz}}+\frac C{1-e^{-(2n-1)\pi q-2\pi qz}};
\end{align}thus the quotient function can be written as
\begin{align}&e^{-2a\pi qz-a^2\pi q}\Bigg[D(a,q)+\sum_{k\ge1}\bigg(\frac{E_k}{1-e^{-(2k-1)\pi q+2\pi qz}}+\frac{F_k}{1-e^{-(2k-1)\pi q-2\pi qz}}\bigg)\Bigg]\\
&=e^{-2a\pi qz-a^2\pi q}\Bigg[D(a,q)+\sum_{k\ge1}\bigg(E_k+\frac{E_k}{e^{(2k-1)\pi q-2\pi qz}-1}+F_k+\frac{F_k}{e^{(2k-1)\pi q+2\pi qz}-1}\bigg)\Bigg]\\
&=e^{-2a\pi qz-a^2\pi q}\Bigg[D_1(a,q)+\sum_{k\ge1}\bigg(\frac{E_k}{e^{(2k-1)\pi q-2\pi qz}-1}+\frac{F_k}{e^{(2k-1)\pi q+2\pi qz}-1}\bigg)\Bigg].
\end{align}Now I need to consult a functional identity for $\phi(z,q)$:$$\phi(z+1,q)=-\phi(z,q)$$(it can be derived from $\phi$'s series definition).
So \begin{align}\frac{\phi(z+1+a,q)}{\phi(z+1,q)}&=e^{-2a\pi q(z+1)-a^2\pi q}\Bigg[D_1(a,q)+\sum_{k\ge1}\bigg(\frac{E_k}{e^{(2k-1)\pi q-2\pi q(z+1)}-1}+\frac{F_k}{e^{(2k-1)\pi q+2\pi q(z+1)}-1}\bigg)\Bigg]\\[2ex]
&=e^{-2a\pi qz-a^2\pi q-2a\pi q}\Bigg[D_1(a,q)+\sum_{k\ge1}\bigg(\frac{E_k}{e^{(2k-3)\pi q-2\pi qz}-1}+\frac{F_k}{e^{(2k+1)\pi q+2\pi qz}-1}\bigg)\Bigg]\\[2ex]
&=e^{-2a\pi qz-a^2\pi q-2a\pi q}\Bigg[D_1(a,q)+\frac{E_1}{e^{-\pi q-2\pi qz}-1}+\frac{E_2}{e^{\pi q-2\pi qz}-1}\\
&\quad+\sum_{k\ge3}\bigg(\frac{E_k}{e^{(2k-3)\pi q-2\pi qz}-1}+\frac{F_{k-1}}{e^{(2k-1)\pi q+2\pi qz}-1}\bigg)\Bigg]\\[2ex]
&=e^{-2a\pi qz-a^2\pi q-2a\pi q}\Bigg[D_1(a,q)-\frac{E_1}{e^{\pi q+2\pi qz}-1}-E_1+\frac{E_2}{e^{\pi q-2\pi qz}-1}\\
&\quad+\sum_{k\ge2}\bigg(\frac{E_{k+1}}{e^{(2k-1)\pi q-2\pi qz}-1}+\frac{F_{k-1}}{e^{(2k-1)\pi q+2\pi qz}-1}\bigg)\Bigg]
\end{align}
and since $\frac{\phi(z+1+a,q)}{\phi(z+1,q)}=\frac{\phi(z+a,q)}{\phi(z,q)}$, you can equate terms with like denominators:$$D_1(a,q)=e^{-2a\pi q}(D_1(a,q)-E_1)$$
$$E_1=e^{-2a\pi q}E_2$$
$$F_1=-e^{-2a\pi q}E_1$$
$$E_{k\ge2}=e^{-2a\pi q}E_{k+1}$$
$$F_{k\ge2}=e^{-2a\pi q}F_{k-1}$$
So now\begin{align}e^{a^2\pi q+2a\pi qz}\frac{\phi(z+a,q)}{\phi(z,q)}&=\frac{E_1}{1-e^{2a\pi q}}+\sum_{k\ge1}\bigg(\frac{e^{(2k-2)a\pi q}E_1}{e^{(2k-1)\pi q-2\pi qz}-1}+\frac{e^{-(2k-2)a\pi q}\cdot-e^{-2a\pi q}E_1}{e^{(2k-1)\pi q+2\pi qz}-1}\bigg)\\
&=\frac{E_1}{1-e^{2a\pi q}}+E_1\sum_{k\ge1}\bigg(\frac{e^{(2k-2)a\pi q}}{e^{(2k-1)\pi q-2\pi qz}-1}-\frac{e^{-2ak\pi q}}{e^{(2k-1)\pi q+2\pi qz}-1}\bigg)
\end{align}and $E_1$ requires only the calculation of one residue:
\begin{align}\lim_{z\to1/2}e^{a^2\pi q+2a\pi qz}(e^{\pi q-2\pi qz}-1)\frac{\phi(z+a,q)}{\phi(z,q)}&=e^{a^2\pi q+a\pi q}\cdot-\varphi(a,q)\frac{-2\pi qe^{\pi q-\pi q}}{\phi'(1/2,q)}\\
&=-2\pi qe^{a^2\pi q+a\pi q}\frac{\varphi(a,q)}{\varphi'(0,q)}
\end{align}After some minor simplifications, we have$$e^{2a\pi qz}\frac{\phi(z+a,q)}{\phi(z,q)}=\frac{\pi q\varphi(a,q)}{\varphi'(0,q)}\Bigg[\operatorname{csch}a\pi q-2\sum_{k\ge1}\bigg(\frac{e^{(2k-1)a\pi q}}{e^{(2k-1)\pi q-2\pi qz}-1}-\frac{e^{(1-2k)a\pi q}}{e^{(2k-1)\pi q+2\pi qz}-1}\bigg)\Bigg]$$
Now I'm going to write the summand as geometric series: I get$$e^{(2k-1)a\pi q}\sum_{n\ge1}e^{-(2k-1)n\pi q+2n\pi qz}-e^{-(2k-1)a\pi q}\sum_{n\ge1}e^{-(2k-1)n\pi q-2n\pi qz}$$Then I swap the indices and sum over $k$:\begin{align}e^{2a\pi qz}\frac{\phi(z+a,q)}{\phi(z,q)}&=\frac{\pi q\varphi(a,q)}{\varphi'(0,q)}\Bigg[\operatorname{csch}a\pi q\\
&\quad-2\sum_{n\ge1}\bigg(e^{(-a+n)\pi q+2n\pi qz}\frac{e^{2\pi q(a-n)}}{1-e^{2\pi q(a-n)}}-e^{(a+n)\pi q-2n\pi qz}\frac{e^{2\pi q(-a-n)}}{1-e^{2\pi q(-a-n)}}\bigg)\Bigg]\\[3ex]
&=\frac{\pi q\varphi(a,q)}{\varphi'(0,q)}\Bigg[\operatorname{csch}a\pi q+\sum_{n\ge1}\bigg(\frac{e^{2n\pi qz}}{\sinh \pi q(a-n)}+\frac{e^{-2n\pi qz}}{\sinh\pi q(a+n)}\bigg)\Bigg]\tag{$\square$}
\end{align}
