It's best to split this up into two cases.

Case 1: $\chi(a) = 1$. Then for $\Re(s) > 1$,
$$\prod_{p \equiv a \pmod{q}} \left(1 - \frac{\chi(p)}{p^s}\right)^{-1} = \prod_{p \equiv a \pmod{q}} \left(1 - \frac{1}{p^s}\right)^{-1} = \sum_{n \in \left\langle \mathcal{P} \right\rangle} \frac{1}{n^s},$$
where $\left\langle \mathcal{P} \right\rangle$ is the multiplicative semigroup generated by the set of primes $\mathcal{P}$ consisting of all $p \equiv a \pmod{q}$. So this is just the Burgess zeta function $\zeta_{\mathcal{P}}(s)$. Now there exist Burgess zeta functions that cannot be holomorphically extended to $1 + it$ for any $t \in \mathbb{R}$ (this is mentioned for example in Terry Tao's paper "A Remark on Partial Sums Involving the Mobius Functions", which I'm pretty sure is available somewhere on the arxiv). In this case, however, I have no idea; perhaps some of the relevant literature discusses it.

Case 2: $\chi(a) = -1$. Then
$$\prod_{p \equiv a \pmod{q}} \left(1 - \frac{\chi(p)}{p^s}\right)^{-1} = \prod_{p \equiv a \pmod{q}} \left(1 + \frac{1}{p^s}\right)^{-1} = \sum_{n \in \left\langle \mathcal{P} \right\rangle} \frac{\lambda(n)}{n^s},$$
where $\lambda(n)$ is the Liouville function. Equivalently,
$$\prod_{p \equiv a \pmod{q}} \left(1 - \frac{\chi(p)}{p^s}\right)^{-1} = \frac{\zeta_{\mathcal{P}}(2s)}{\zeta_{\mathcal{P}}(s)},$$
so it comes down to the same thing; determining whether $\zeta_{\mathcal{P}}(s)$ extends holomorphically to the line $\Re(s) = 1$ and beyond.

EDIT: Recall that the prime number theorem for arithmetic progressions says that
$$\pi(x;q,a) = \frac{1}{\varphi(q)} \mathrm{li}(x) + O_A(x \exp(-c_1 (\log x)^{1/2})$$
for fixed $A > 0$ with $q \leq (\log x)^A$. An application of a result of Diamond (cf. *Asymptotic Distribution of Beurling's Generalized Integers*) then implies that
$$N_{\mathcal{P}}(x) = \sum_{n \in \left\langle \mathcal{P} \right\rangle, \; n \leq x}{1} = a x + O_A(x \exp(-c_2 (\log x \log \log x)^{1/3})$$
for some particular $a > 0$. By partial summation, we have that for $\Re(s) > 1$,
$$\zeta_{\mathcal{P}}(s) = \frac{as}{s-1} + s \int_{1}^{\infty} \frac{N_{\mathcal{P}}(x) - ax}{x^{s+1}} \: dx .$$
Diamond's result implies that this integral is uniformly convergent for $\Re(s) \geq 1$, and so it is continuous in this half-plane. Thus $\zeta_{\mathcal{P}}(s) = c/(s-1) + r_0(s)$ with $r_0(s)$ continuous for $\Re(s) \geq 1$, and so $\zeta_{\mathcal{P}}(s)$ extends to $\Re(s) \geq 1$ with a singularity at $s = 1$. Moreover, it is not difficult to show that $\zeta_{\mathcal{P}}(1+it) \neq 0$ for all $t \in \mathbb{R}$; a version of this is shown in Montgomery and Vaughan's *Multiplicative Number Theory I: Classical Theory* section 8.4.

Note also that assuming the generalised Riemann Hypothesis, it is possible to strengthen this meromorphic extension of $\zeta_{\mathcal{P}}(s)$ to $\Re(s) > 1/2$ with $\zeta_{\mathcal{P}}(s)$ nonvanishing in this open half-plane; see Titus W. Hilberdink and Michel L. Lapidus, *Beurling Zeta Functions, Generalised Primes and Fractal Membranes*.