I have tried to prove the followng.
If $f$ satisfies the Holder condition with exponent $\alpha\in(0,1]$ in $[-\pi,\pi]$, then $c_n(f)=\mathcal{O}(\vert n \vert ^{-\alpha})$ as $n\rightarrow\infty$ where $$ c_n(f)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx. $$
At first, I assumed that $f$ is a periodic function and proved this statement as follows.
$$2\pi c_n(f)=\int_{-\pi}^{\pi} f(y)e^{-iny}dy=\int_{-\pi+\pi/n}^{\pi+\pi/n} f(y)e^{-iny}dy=e^{-i\pi}\int_{-\pi}^{\pi}f(x+\pi/n)e^{-inx}dx$$
So we have
$$ \begin{align} \vert -4\pi c_n(f) \vert &= \left\vert \int_{-\pi}^{\pi}(f(x+\pi/n)-f(x))e^{-inx}dx \right\vert \newline &\leq 2\pi\sup_{x,x+\pi/n\in[-\pi,\pi]}\left\vert f(x+\pi/n)-f(x) \right\vert \newline &\leq C(\pi/\vert n \vert)^\alpha \end{align} $$
But if $f$ is not periodic, then how can I prove the above statement? At least if $f$ is defined on any interval $[a,b]$ and if $f(a)=f(b)$ then we can extend $f$ as a periodic function and there exists another periodic function $g$ defined on $[-\pi,\pi]$ whose Fourier coefficients are same as those of $f$; so this is not the case.
I would like to post my own answer to this question since I fully resolve this issue.
In order to factor out $n^{-\alpha}$, we express the integral as a sum of finite $n>0$ integrals as follows: \begin{align} c_n(f) &= \frac{1}{b-a}\sum_{j=0}^{n-1}\int_{a+\frac{(b-a)j}{n}}^{a+\frac{(b-a)(j+1)}{n}} f(x) e^{inx} dx \\ &= \frac{1}{b-a}\sum_{j=0}^{n-1}\int_{a+\frac{(b-a)j}{n}}^{a+\frac{(b-a)(j+1)}{n}} \left(f(x) - f\left(a+\frac{(b-a)j}{n}\right)\right) e^{inx} dx \end{align} This can be still done as $n\rightarrow\infty$ by Riemann-Lebesgue lemma. Then we have \begin{align} \vert c_n(f) \vert &\leq \frac{1}{b-a}\sum_{j=0}^{n-1}\int_{a+\frac{(b-a)j}{n}}^{a+\frac{(b-a)(j+1)}{n}} \left| f(x) - f\left(a+\frac{(b-a)j}{n}\right) \right| dx \\ &\leq \frac{1}{b-a}\sum_{j=0}^{n-1}\frac{b-a}{n}C\left|\frac{b-a}{n}\right|^\alpha \\ &\leq C|n|^{-\alpha} \end{align} using the Holder condition of $f$.