No, you cannot put any better bound than S_{N} = *o*(N). There is a general technique, using the Baire category theorem of proving the existence of counterexamples to problems like this (which I discovered while trying to find a counterexample to a question by David Speyer, link). I see that Helge's answer is also pointing towards the same result.

First, for θ irrational,
$$
S_N/N=\frac{1}{N}\sum_{n=1}^N1_{\{0< n\theta/2 <1/2{\rm\ (mod\ 1)}\}}-\frac{1}{N}\sum_{n=1}^N1_{\{1/2< n\theta/2 <1{\rm\ (mod\ 1)}\}}
$$
By Weyl's equidistribution theorem, both sides on the right hand side tend to 1/2 and S_{N} / N → 0, so S_{N} = *o*(N).

It is not possible to do better than this. In fact, if f: ℕ → ℝ^{+} is any function satisfying liminf f(N) / N = 0 then there will be an uncountable dense set of irrational θ for which limsup S_{N} / f(N) = ∞. In particular, using f(n) = n^{x} for x < 1 rules out bounds such as S_{n} = *O*(n^{x}).
In fact, we can find a set of such θ as an intersection of countably many open dense subsets of ℝ, so the Baire category theorem shows the existence of uncountably many counterexamples.

Let u(x) = 1_{{0≤[x/2]<1/2}} - 1_{{1/2≤[x/2]<1}} where [x] is the fractional part of x, and S_{N}(θ) = Σ_{n≤N} u(nθ). Let U_{K} be the set
$$
U_K=\left\{\theta\in\mathbb{R}\colon S_n(\theta)>Kf(n){\rm\ for\ some\ }n\ge K\right\}.
$$
This contains a dense open subset of ℝ. In fact, if θ = 2p/q for q odd then, for 1 ≤ n < q, u((q-n)θ) = -u(nθ). So, S_{q-1}(θ) = 0 and S_{q}(θ) = 1. Then, by periodicity of [nθ/2], S_{nq} (θ) = n and S_{n}(θ) increases linearly. So, S_{n}(θ) > Kf(n) for infinitely many n, and θ ∈ U_{K}. By right continuity of u, (θ,θ+ε) ⊆ U_{K} for small enough ε. This shows that (2p/q,2p/q+ε) is contained in the interior of U_{K} and, as such 2p/q are dense, the interior of U_{K} is a dense open subset of ℝ. The Baire category theorem implies that
$$
U\equiv\bigcap_{K=1}^\infty U_K
$$
is an uncountable dense subset of ℝ and, by construction, for any θ ∈ U, limsup S_{n}(θ) / f(n) > K for each K.

The further question was asked in the comment: are there *any* irrational θ for which S_{N} = *O*(N^{x}) for x < 1. The answer is yes. In fact this holds for almost every θ and every x > 1/2.

The idea is to consider rational approximations to θ, |θ/2 - p/q| ≤ q^{-2}. Then, there will be an integer 1 ≤ a < q such that |1/2 - [ap/q]| ≤ 1/(2q). So, |1/2-[aθ/2]| ≤ 1/q. With u() as above, it follows that u(nθ) + u((n+a)θ) = 0 unless -2/q ≤ nθ ≤ 2/q (mod 1). So, there is a lot of cancellation in S_{N}(θ),

$$
\begin{array}
\displaystyle
\vert S_N(\theta)\vert &\displaystyle \le a +\sum_{n=1}^N1_{\{-2/q\le n\theta\le 2/q{\rm\ (mod\ 1)}\}}\\\\
&\displaystyle\le 2q +\sum_{n=0}^{\lfloor N/q\rfloor}\sum_{m=1}^q1_{\{-2/q\le nq\theta+m\theta\le 2/q{\rm\ (mod\ 1}\}}\\\\
&\displaystyle\le 2q+\sum_{n=0}^{\lfloor N/q\rfloor}\sum_{m=1}^q1_{\{-4/q\le nq\theta+2mp/q\le 4/q{\rm\ (mod\ 1)}\}}
\end{array}
$$
The points 2mp/q (mod 1) are equally spaced. If q is odd then they have spacing 1/q and no more than 9 of them can lie in an interval of length 8/q. If q is even then the spacing is 2/q and no more than 5 can lie in such an interval. In either case, the final sum over m above is bounded by 10=5*2.
$$
\vert S_N(\theta)\vert\le 2q+10N/q.
$$
If θ has irrationality measure less than α then, for large enough N, the rational approximation p/q can be chosen such that N^{1/2} ≤ q ≤N^{(α-1)/2},
$$
\vert S_N(\theta)\vert\le 2N^{(\alpha-1)/2}+10N^{1/2}.
$$
In particular, if θ has irrationality measure 2 then $S_N=O(N^x)$ for every $x>1/2$. But, almost every real number has irrationality measure 2.