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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 [Edit: I think. There is a mistake in the proof below, but hopefully it can be fixed], 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 m² ≡ n² (mod b), so θm² - θn² 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_bN(θ) = NS_b(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_N(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 $$ \begin{array} {}\displaystyle\theta=\sum_{n=1}^\infty\frac{1}{p_n}&&(*) \end{array} $$ 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_n grows quickly, such as p_n ≥ 2ⁿ. 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, θ will be irrational (transcendental, in fact).

Now, let us choose p₁,p₂,... inductively. Suppose that odd primes p₁,...,p_m have already been chosen, and set $\theta_m=1/p_1+\cdots+1/p_m$. [Edit: the following line is wrong. θ_m is not in the form for which (3) applies. I think this proof can probably be rescued. I'll have to come back to it] By (3), $S_N(\theta_m)/N$ converges to a nonzero limit, so we can choose an $N_m$ with $\vert S_{N_m}(\theta_m)/(h(N_m)N_m)\vert>m$. 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 θ 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 N₁,N₂,... of integers such that $\vert S_{N_m}(\theta)/{(h(N_m)N_m)}\vert\ge m$, so $\sup_N\vert S_N(\theta)/(h(N)N)\vert=\infty$.