No, you cannot put any better bound than S<sub>N</sub> = *o*(N). There is a general technique, using the [Baire category theorem][1] 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][2]). I see that Helge's answer is also pointing towards the same result.
First, for θ irrational,
$$
S_N/N=\frac{1}{N}\sum_{n=1}^N1_{\{0< n\theta <1/2{\rm\ (mod\ 1)}\}}-\frac{1}{N}\sum_{n=1}^N1_{\{1/2< n\theta <1{\rm\ (mod\ 1)}\}}
$$
By [Weyl's equidistribution theorem][3], both sides on the right hand side tend to 1/2 and S<sub>N</sub> / N → 0, so S<sub>N</sub> = *o*(N).
It is not possible to do better than this. In fact, if f: ℕ → ℝ<sup>+</sup> is any function satisfying liminf f(N) / N = 0 then there will be an uncountable dense set of irrational θ for which limsup S<sub>N</sub> / f(N) = ∞. In particular, using f(n) = n<sup>x</sup> for x < 1 rules out bounds such as S<sub>n</sub> = *O*(n<sup>x</sup>).
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<sub>{0≤[x]<1/2}</sub> - 1<sub>{1/2≤[x]<1}</sub> where [x] is the fractional part of x, and S<sub>N</sub>(θ) = Σ<sub>n≤N</sub> u(nθ). Let U<sub>K</sub> 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 = p/q for q odd then, for 1 ≤ n < q, u((n-q)θ) = -u(qθ). So, S<sub>q-1</sub>(θ) = 0 and S<sub>q</sub>(θ) = 1. Then, by periodicity of [nθ], S<sub>nq</sub> (θ) = n and S<sub>n</sub>(θ) increases linearly. So, S<sub>n</sub>(θ) > Kf(n) for infinitely many n, and θ ∈ U<sub>K</sub>. By right continuity of u, (θ,θ+ε) ⊆ U<sub>K</sub> for small enough ε. This shows that (p/q,p/q+ε) is contained in the interior of U<sub>K</sub> and, as such p/q are dense, the interior of U<sub>K</sub> 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<sub>n</sub>(θ) / f(n) > K for each K.
[1]: http://en.wikipedia.org/w/index.php?title=Baire_category_theorem&oldid=379733059
[2]: http://mathoverflow.net/questions/35902/does-weyls-inequality-prove-equidistribution/35976#35976
[3]: http://en.wikipedia.org/wiki/Weyl%2527s_equidistribution_theorem