It looks like the conjecture is (almost) true -- $\zeta_\theta(s)$ extends to a meromorphic function on $\mathbb{C}$ for all real quadratic irrationals $\theta$ with poles at $s=2$, $s=1$ and in the infinite set of vertical lines $\lbrace -2n+i\mathbb{R}\colon n\in\mathbb{Z}_{\ge0}\rbrace$. I'm not sure about the more tentative conjecture for arbitrary algebraic numbers. Also, more simply, it will extend to a meromorphic function over $\Re(s) > 0$ whenever $\theta$ cannot be approximated too well by rationals (in particular, if $\theta$ has irrationality measure 2 which, by the Thue-Siegel-Roth theorem, includes all algebraic irrationals). Conversely, if an irrational $\theta$ can be approximated too well by rationals then there will not be a meromorphic extension, which will be the case outside of a meagre subset of $\mathbb{R}$.
[Edit: As Pieter pointed out in the comments, my initial answer carelessely missed out a rather important term, so was not correct. I've updated the proof sketch to deal with this term, and restricted to quadratic irrationals for the meromorphic extension. This is very sketchy, and I only go through the full details for $\mathbb{Z}+\mathbb{Z}\theta$ a quadratic number field, but I think it does generalize to arbitrary quadratic irrationals in a similar way.]
First, let $\lbrace x\rbrace = x - \lfloor x\rfloor$ denote the fractional part of $x$, and write $$ \tilde\zeta_\theta(s)=\sum_{n\ge1}\frac{\lbrace n\theta\rbrace-1/2}{n^s}, $$ which converges, at least, on $\Re(s) > 1$. Then, the function $\zeta_\theta$ defined in the question is $$ \zeta_\theta(s)=\theta\zeta(s-1)-\frac12\zeta(s)-\tilde\zeta_\theta(s), $$ where $\zeta$ is the Riemann zeta function. So we have an extension to $\Re(s) > 1$. In the case where $\theta$ is rational then $\lbrace n\theta\rbrace-1/2$ is periodic in $n$, so $\tilde\zeta_\theta$ is a linear combination of Dirichlet L-functions and has a meromorphic extension to $\mathbb{C}$. Henceforth, I will only consider irrational $\theta$. Setting $F(x)=\sum_{1\le n\le x}(\lbrace n\theta\rbrace-1/2)$, the equidistribution theorem states that $F(x)/x\to0$ as $x\to\infty$. Rearranginging the expression for $\tilde\zeta_\theta$ gives $$ \tilde\zeta_\theta(s)=\sum_{n\ge1}\frac{F(n)}{n}n^{1-s}\left(1-(1+1/n)^{-s}\right) $$ If $\lvert F(n)/n\rvert$ is bounded by some $\delta > 0$ then this expression will be bounded by $\delta\zeta(s)$ for all $s > 1$. So, by equidistribution, it follows that $(s-1)\tilde\zeta_\theta(s)\to0$ as $s\to1$ (on $s > 1$). If a meromorphic extension existed in a neighbourhood of 1 then this would imply that $\tilde\zeta_\theta$ is bounded near 1. However, if $\theta=p/q$ is rational ($p,q$ coprime) then $F(x)/x\to-1/(2q)$ so, if $x$ is too closely approximated by rationals then there will be a slight bias in $F(x)/x$ and $\tilde\zeta_\theta(s)$ will not be bounded close to 1. I think that the existence of rational approximations $p_n/q_n$ with $q_n^2\log\lvert\theta-p_n/q_n\rvert\to-\infty$ is enough for this to happen.
Now, suppose that $\theta$ has irrationality measure no greater than some finite $\gamma$, so that there are only finitely many rational approximations $\lvert\theta-p/q\rvert\le q^{-\gamma-\delta}$ for any $\delta > 0$. Equivalently, there is a constant $C_\delta > 0$ such that $n^{\gamma-1+\delta}\lvert n\theta-m\rvert\ge C$ for all positive integer $n$ and integer $m$. Then, the discrepancy of the set $\left\lbrace \lbrace k\theta\rbrace\colon 1\le k\le n\right\rbrace$ is bounded by $Kn^{-1/(\gamma-1)+\delta}$ (for a constant $K$). This means that $F(n)/n$ is bounded by $2Kn^{-1/(\gamma-1)+\delta}$ and, hence, the expression above for $\tilde\zeta_\theta(s)$ has summand going to zero at rate $O(n^{-s-1/(\gamma-1)+\delta})$. So, $\tilde\zeta_\theta(s)$ is an analytic function on $\Re(s) > (\gamma-2)/(\gamma-1)$. In particular, for algebraic irrationals, $\gamma=2$ and we get a meromorphic extension of $\zeta_\theta(s)$ to $\Re(s) > 0$ with simple poles at $s=2$ and $s=1$.
I now attempt to extend to the whole of $\mathbb{C}$ via Hurwitz's formula which, for $\Re(s) > 0$ and irrational $x$ in the unit interval, gives $$ \begin{array}{l} &\sum_{n\ge1}e^{2\pi inx}n^{-s}=(2\pi)^{s-1}\Gamma(1-s)\left(e^{i\pi(1-s)/2}\zeta(1-s,x)+e^{i\pi(s-1)/2}\zeta(1-s,1-x)\right) \end{array} $$ where $\zeta(s,x)=\sum_{n\ge0}(n+x)^{-s}$ is the Hurwitz Zeta function. Plugging in the Fourier series $\lbrace x\rbrace=1/2+\sum_{k\not=0}(i/(2\pi k))e^{2\pi ikx}$, applying Hurwitz's formula, and being careless about when the sums converge/commute (it can be made more rigorous) gives, $$ \begin{array}{rl} \tilde\zeta_\theta(s)&=\sum_{n\ge1}\sum_{k\not=0}\frac{ie^{2\pi ink\theta}}{2\pi kn^s}\cr &=(2\pi)^{s-2}\Gamma(1-s)\sum_{k\not=0}\frac{i}{k}\left(e^{i\pi(1-s)/2}\zeta(1-s,\lbrace k\theta\rbrace)+e^{i\pi(s-1)/2}\zeta(1-s,1-\lbrace k\theta\rbrace)\right)\cr &=(2\pi)^{s-2}\Gamma(1-s)\sum_{k\not=0}\frac{i}{k}\left(e^{i\pi(1-s)/2}\zeta(1-s,\lbrace k\theta\rbrace)-e^{i\pi(s-1)/2}\zeta(1-s,\lbrace k\theta\rbrace)\right)\cr &=2\sin((s-1)\pi/2)(2\pi)^{s-2}\Gamma(1-s)\sum_{k\not=0}k^{-1}\zeta(1-s,\lbrace k\theta\rbrace)\cr &=2\sin((s-1)\pi/2)(2\pi)^{s-2}\Gamma(1-s)\sum_{n\ge0}\sum_{k\not=0}k^{-1}(n+\lbrace k\theta\rbrace)^{s-1}. \end{array} $$ Split up the summation as $$ \sum_{n\ge1}n^{s-1}\sum_{k\not=0}k^{-1}\left(1+\lbrace k\theta\rbrace/n\right)^{s-1}+\sum_{k\not=0}k^{-1}\lbrace k\theta\rbrace^{s-1}. $$ The first term can be handled easily. Use the fact that $\theta$ has irrationality measure 2, so that $\left\lbrace\lbrace k\theta\rbrace\colon k=1,\ldots,n\right\rbrace$ has discrepancy $O(n^{-1+\delta})$. As the function $x\mapsto(1+x/n)^{s-1}$ has bounded variation of size $O(1/n)$ over the unit interval, the summation over $k$ converges and is of size $O(1/n)$. Therefore, the sum over $n$ converges on $\Re(s) < 1$, showing that the first term converges to an analytic function on $\Re(s) < 1$.
The final summation over $k$ is more problematic. Assuming $\theta$ is a quadratic irrational, and using $\bar z$ to denote the conjugate of $z\in\mathbb{Q}(\theta)$, it can be rewritten (up to order of summation) as $$ (\bar\theta-\theta)\sum_{0 < z < 1}\frac{z^{s-1}}{\bar z-z}=(\bar\theta-\theta)\sum_{j=0}^\infty\sum_{0 < z < 1}{\bar z}^{-j-1}z^{s+j-1}. $$ Here, I have substituted in $z=\lbrace k\theta\rbrace$, and the summation is over all $z\in\mathbb{Z}+\mathbb{Z}\theta$ lying in the unit interval. For each bounded region for $s$ on which we want to extend. We only actually have to expand out a finite number of terms in the sum over $j$ above, up until $\Re(s)+j-1$ is positive. In that case, the numerator of the remainder is $z^{s+j-1}=\lbrace k\theta\rbrace^{s+j-1}$. As $x^{s+j-1}$ will have finite variation over the unit interval, the remainder term will converge to an analytic function. So, we just need to show that each term in the sum over $j$ extends to a meromorphic function. I'll now restrict to the case where $\mathbb{Z}+\mathbb{Z}\theta$ is a number field. For example, $\theta=\sqrt{d}$ for squarefree $d\not\cong1$ (mod 4) or $\theta=(1+\sqrt{d})/2$ for squarefree $d\cong1$ (mod 4). I think the general quadratic case follows in a similar manner, but it gets a bit messier. Let $\eta$ be the fundamental unit of $\mathbb{Z}[\theta]$ lying in the unit interval. Then, each $z$ in the unit interval can be written uniquely as $\eta^ny$ for $y$ in $[\eta,1)$ and nonnegative integer $n$. Writing $\epsilon=\bar\eta\eta=\pm1$, the term for a fixed $j$ in the expansion above is (again, being a bit careless about order of summation), $$ \begin{array}{rl} \sum_{0 < z < 1}{\bar z}^{-j-1}z^{s+j-1}&=\sum_{\eta\le z < 1}\sum_{n=0}^{\infty}(\bar\eta ^n\bar z)^{-j-1}(\eta^nz)^{s+j-1}\cr &=\sum_{\eta\le z\le1}\sum_{n=0}^\infty{\bar z}^{-j-1}z^{s-j-1}\epsilon^{n(j+1)}\eta^{n(s+2j)}\cr &=\left(1-\epsilon^{j+1}\eta^{s+2j}\right)^{-1}\sum_{\eta\le z < 1}{\bar z}^{-j-1}z^{s-j-1}. \end{array} $$ Now that the summation has being restricted to $z\ge\eta$, the term $z^{s-j-1}$ will be bounded, and the sum is guarateed to converge. To see this, substitute back $z=\lbrace k\theta\rbrace$ and $\bar z=k(\bar\theta-\theta)+\lbrace k\theta\rbrace$. The summand is of size $O(k^{-1-j})$, so the sum converges absolutely for $j\ge1$. For $j=0$ it is $(\bar\theta-\theta)^{-1}\lbrace k\theta\rbrace^{s-1}/k + O(k^{-2})$ for $\lbrace k\theta\rbrace\ge\eta$. As $x^{s-1}$ has finite variation over the interval $[\eta,1)$, the sum for $j=0$ also converges.
Finally, the term $(1-\epsilon^{j+1}\eta^{s+2j})^{-1}$ is meromorphic on $\mathbb{C}$ with zeros on the vertical line $-2j+i\mathbb{R}$. So, $\zeta_\theta$ does extend with poles in the claimed set.