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.
I don't think that it is 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 $S_N/N=O(N^{-1})$). [Edit: There is an error below, but I'm hopeful it can be fixed]. 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, SbN(θ) = NSb(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/b for coprime integers a,b with b odd and φ(b)≥b/2 (φ is Euler's totient function. Then $x=S_b(\theta)/b$ is nonzero and, by (2), $S_N(\theta)/N$ tends to a nonzero limit.
Proof: As a,b are coprime, $u=e^{2\pi i \theta}$ is a primitive b'th root of unity, with minimal polynomial of degree φ(b)≥b/2 over the rationals. So $$ S_b(\theta)=\sum_{k=1}^{b}u^{k^2}=1+2\sum_{k=1}^{(b-1)/2}u^{k^2} $$ is nonzero. [Edit: There is still an error here. I need to show that $\{u^{k^2}\colon k=0,\ldots,(b-1)/2\}$ is linearly independent over the rationals, which I haven't done. Hopefully this can be fixed]
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 pn grows quickly, such as pn ≥ 2n. Now, let us choose p1,p2,... inductively. Suppose that odd primes p1,...,pm have already been chosen, and set $\theta_m=1/p_1+\cdots+1/p_m$. We can write θm=am/bm where bm=p1...pm. If we make sure that $p_m>2^{m-n}b_n^n$ for all m>n, this will give the following rational approximations $$ \vert\theta-a_m/b_m\vert=\sum_{k=n+1}^\infty \frac{1}{p_k}\le\frac{1}{p_n}^n $$ and, by Liouville's theorem, θ will be irrational (transcendental, in fact).
Also, assuming that pn are chosen large enough that $\prod_n(1-1/p_n)>1/2$ then we have φ(bm) ≥ bm/2 and (3) can be applied. 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 $\sup_N\vert S_N(\theta)/(h(N)N)\vert=\infty$.