This response is in answer to David's further question about whether it is possible to bound the rate at which $S_N/N$ tends to zero, as he was wanting to use Weyl's inequality to do.
In fact it is not possible, even in the case d=2 and $f(n)=\theta n^2$. (for d=1 it is not hard to show that $S_N$ is bounded so $S_N/N=O(N^{-1})$). I'll write
$$
S_N(\theta)=\sum_{n=1}^Ne^{2\pi i\theta n^2}
$$
in the following. Then, I'll try to show that, for any given h(N)->0, it is not true that $S_N(\theta)/N$ always tends to zero at rate O(h(N)).
> (1) Let $h\colon\mathbb{N}\to\\mathbb{R}_+$ satisfy $\lim_nh(n)=0$. Then, there exists an irrational $\theta$ such that $\sup_N\vert S_N(\theta)/(h(N)N)\vert=\infty$
I'll construct this by choosing θ as the limit of a very quickly converging sequence of rational numbers. Let's break up the construction, starting with the case where θ is actually rational.
> (2) Let θ = a/b for integers a,b with b > 0. Setting $x=S_b(\theta)/b$ then $S_N(\theta)/N\to x$ as $N\to\infty$.
*Proof:*
If m ≡ n (mod b) then m<sup>2</sup> ≡ n<sup>2</sup> (mod b), so θm<sup>2</sup> - θn<sup>2</sup> is an integer, and $e^{2\pi i\theta m^2}=e^{2\pi i \theta n^2}$. This shows that $n\mapsto e^{2\pi i\theta n^2}$ has period b.
This gives
$$
S_{bN}(\theta)=\sum_{j=0}^{N-1}\sum_{k=1}^{b}e^{2\pi i\theta(jb+k)^2}=N\sum_{k=1}^be^{2\pi i\theta k^2}.
$$
So, S<sub>bN</sub>(θ) = NS<sub>b</sub>(θ). Now, any N can be written as N = bM + R for some R < b. Then, $\vert S_N-MS_b\vert\le R$ and, dividing by N gives $\vert S_N/N-S_b/b\vert\to0$ as N goes to infinity.
> (3) Let θ = a/p for integer a and odd prime p. Then $x=S_p(\theta)/p$ is nonzero and, by (2), $S_N(\theta)/N$ tends to a nonzero limit.
*Proof:*
Assume without loss of generality that p does not divide a (otherwise, S<sub>N</sub>(θ) is strictly positive). Then $u=e^{2\pi i \theta}$ is a primitive p'th root of unity, with minimal polynomial $X^{p-1}+X^{p-2}+\cdots+X+1$ over the rationals. It follows that no proper subset of $\{1,u,u^2,\ldots,u^{p-1}\}$ is linearly dependent over the rationals. So
$$
S_p(\theta)=\sum_{k=1}^{p}u^{k^2}=1+2\sum_{k=1}^{(p-1)/2}u^{k^2}
$$
is nonzero.
Using this, a θ can be constructed proving (1) above.
> Let $h\colon\mathbb{N}\to\mathbb{R}$ satisfy $\lim_nh(n)=0$. Then, there exists a sequence $p_k$ of prime numbers, tending to infinity, such that the sum
$$
\begin{array}
{}\displaystyle\theta=\sum_n\frac{1}{p_n}&&(*)
\end{array}
$$
converge to an irrational number, and $\sup_NS_N(\theta)/(h(N)N)=\infty$.
*Proof:*
In order for the sum in (*) to converge, we have to require that p<sub>n</sub> grows quickly, such as p<sub>n</sub> ≥ 2<sup>n</sup>. In fact, choosing $p_{n+1}\ge 2p_n^n$ gives the following rational approximations,
$$
\vert\theta-a/p_n\vert=\sum_{k=n+1}^\infty \frac{1}{p_k}\le\frac{1}{p_n^n}
$$
and, by [Liouville's theorem][1], θ will be irrational (transcendental, in fact).
Now, let us choose p<sub>1</sub>,p<sub>2</sub>,... inductively. Suppose that odd primes p<sub>1</sub>,...,p<sub>m</sub> have already been chosen, and set $\theta_m=1/p_1+\cdots+1/p_m$. By (3), $S_N(\theta_m)/N$ converges to a nonzero limit, so we can choose an $N_m$ with $\vert S_{N_m}(\theta_m)/(h(N_m)N_m)>m$.
By continuity, there is an $\epsilon>0$ such that $\vert S_{N_m}(\theta)/{N_m}\vert>m$ whenever $\vert\theta-\theta_m\vert\le\epsilon$. To guarantee that our value of $\theta$ defined by (*) satisfies this, it is only necessary to choose $p_n>2^{n-m}/\epsilon$ for all n > m.
Proceeding in this way, we can choose a quickly increasing sequence of prime numbers, where each choice of prime number imposes a lower bound on the following terms in the sequence. It also provides us with a sequence N<sub>1</sub>,N<sub>2</sub>,... of integers such that $\vert S_{N_m}(\theta)/{N_m}\vert\ge m$, so $\sup_N\vert S_N(\theta)/N\vert=\infty$.
[1]: http://en.wikipedia.org/wiki/Liouville_number