Here is an argument essentially due to fedja I learned about thirteen years ago on artofproblemsolving.com.
Proposition: if $f$ is $2$-periodic Riemann integrable such that $\sup_{n \geq 1} \left|\sum_{k=1}^{n}f(kx)\right|<C<\infty$ for some irrational $x \in \mathbb{R} \setminus \mathbb{Q}$ then necessarily $$ \sum_{m\neq 0}\, \left| \frac{\hat{f}(m)}{e^{i \pi m x}-1}\right|^{2}<C^{2}. \quad (1) $$.
Notice that the proposition solves the question. Indeed, $f(t)=(-1)^{\lfloor{t}\rfloor}$ is 2-periodic with Fourier coefficients $\hat{f}(m) = \left| \frac{1}{2}\int_{0}^{2}f(t)e^{-i\pi m x}dx \right| \approx \frac{1}{m}$ for $m$ odd, and $\hat{f}(m)=0$ for even $m$. There are infinitely many odd numbers $m$ such that $\mathrm{dist}\left(mx, 2\mathbb{Z}\right) < \frac{C'}{m}$ (it does not follow directly from Dirichlet's rational approximation theorem, however, I think it is not difficult to adapt it here or use Minkowski's theorem on product of two linear forms). Therefore, in the left hand side of (1) there are infinitely many terms comparable to 1 so we get a contradiction.
Proof of the proposition:
Let me first assume that $f$ is continuous and then I will explain later what to do in the arbitrary case. Let $S_{m} = \sum_{k=1}^{m-1}f(kx)$. Then the boundedness of $|S_{M+n}-S_{M}|$ implies $$ |\sum_{k=0}^{n-1} f(Mx+kx)|<2C \quad \text{for all} \quad M>1 \quad (2) $$ and all $n\geq 1$. By density of the sequence $\{Mx\, \mathrm{mod}\, (2)\}_{M>1}$ in $[0,2)$ we conclude $$ |\sum_{k=0}^{n-1} f(t+kx)|\leq 2C \quad \text{for all} \quad t \in [0,2). \quad (3) $$ Let us convolve $f$ with Fejer Kernel so that the new function $F$ is now a trigonometric polynomial with almost the same Fourier coefficients, and clearly it also satisfies the inequality (3). After expanding $F$ into its Fourier series $F(s) = \sum_{m} \hat{F}(m) e^{i\pi m s}$ (finite sum), and using (3) for $F$ we obtain $$ \left| \sum_{m} \hat{F}(m)e^{i \pi m t} \, \frac{e^{i \pi m n x} -1}{e^{i \pi m x} -1} \right| <2C $$ In particular its $L^{2}$ norm is bounded, i.e., $$ \sum_{m\neq 0} \left|\hat{F}(m)\, \frac{e^{i \pi m n x} -1}{e^{i \pi m x} -1}\right|^{2} <4C^{2} $$ Invoking the density of the sequence $\{n x\, \mathrm{mod}\, (2)\}_{n \geq 1}$ again we conclude (1) for $F$. Finally it remains to remove the convolution with Fejer kernel to conclude (1) for $f$ (we have nonnegative terms in the sum on the left hand side of (1), so we can just cut the sum and take the limit $\hat{F}(m) \to \hat{f}(m)$)
In general, if $f$ is not continuous, let $K_{N}$ be the Fejer kernel. Split $Mx=M_{1}x+M_{2}x$ in (2), multiply (2) by $K_{N}(M_{1}x)$ and take the average in $M_{1}$ and use the Riemann integral criteria for equidistribution of $\{M_{1} x\, \mathrm{mod}\, (2)\}_{M_{1}\geq 1}$ to conclude $\left| \sum_{k=0}^{n-1} F(M_{2}x + kx)\right|<2C$ and now due to continuity of $F$ and density of $\{M_{2} x\}$ we conclude (3) for $F$ and the rest of the argument proceeds in the same way. $\square$