Is the integral always nonzero? Let 
$$I_{n,p,a}:=\int_0^{\infty-} \frac{g_{n,a}(t)}{t^{p+1}} \, dt,$$
where 
$$(*)\qquad\qquad\qquad n\in\mathbb N,\quad -\infty<a<\infty,\quad p_{n,a} < p < n,\qquad\qquad\qquad\qquad\phantom{x} $$
$p_{n,a}$ equals $0$ or $-1$ depending on whether $a$ belongs to the set $\{2j-n\colon\, j=0,\dots,n\}$ or not, 
and 
$$g_{n,a}(t):=e^{i a t} \left(e^{i t}-e^{-i t}\right)^n.$$
The integral $I_{n,p,a}$ arises in certain explicit expressions of certain linear functionals of a measure in terms of the Fourier transform of the measure. It is not hard to see that for non-integer $p\in(p_{n,a},n)$
$$I_{n,p,a}=i^p\, \Gamma (-p)\, \sum _{j=0}^n (-1)^j \binom{n}{j}\, \overline{(2j-n-a)^p},$$
where $\overline{z}$ denotes the complex conjugate of $z$, and $I_{n,p,a}$ is continuous in $p\in(p_{n,a},n)$, so that the values of $I_{n,p,a}$ in the case when $p$ is an integer in the interval $(p_{n,a},n)$ can be obtained by the l'Hospital rule. 
Here we follow the usual convention that the (principal value of the) argument $\arg z$ of a nonzero complex number $z$ belongs to the interval $(-\pi,\pi]$, and then $z^p$ is understood as $|z|^p e^{ip\arg z}$. 
The question is this: Do there exist $n$, $p$, and $a$ such that conditions $(*)$ hold and $I_{n,p,a}=0$? It appears that the answer is negative. 
Below are pictures of the sets $\big\{\big((1 + a^2)^{(n - p)/2}I_{n,p,\,a},\,6a\big)\colon -10<a<10\big\}$ for $(n,p)=(3,0.4)$ and 
$\big\{\big((1 + a^2)^{(n - p)/2}I_{n,p,a},\,10a\big)\colon -6<a<10\big\}$ for $(n,p)=(4,1.4)$ in $\mathbb{C}\times\mathbb R$. In each picture, shown also is the (dotted) vertical line through the point $0\in\mathbb C$, respectively $\big\{(0,\,6a\big)\colon -10<a<10\big\}$ for $(n,p)=(3,0.4)$ and 
$\big\{(0,\,10a\big)\colon -6<a<10\big\}$ for $(n,p)=(4,1.4)$. The factor $\big((1 + a^2)^{(n - p)/2}$ at $I_{n,p,a}$ is in accordance with the expression for $I_{n,p,a}$ given in the answer in the case when $|a|>n$. 

 A: This is only a very partial answer. Suppose that $a<-n$ and $p$ is a non-integer in the interval $(-1,n)$. Then $2j-n-a>0$ for all $j=0,\dots,n$. So, by the mean-value theorem applied (say, repeatedly) to the $n$-fold symmetric difference in the expression of $I_{n,p,a}$ in $(**)$ in the question statement, one has 
$$I_{n,p,a}=i^p\, \Gamma (-p)\, p(p-1)\cdots(p-n+1)2^n(2j_{n,p,a}-n-a)^{p-n}$$
for some real $j_{n,p,a}\in(0,n)$, so that 
$$i^{-p}(-1)^n I_{n,p,a}=\Gamma (n-p)\,2^n(2j_{n,p,a}-n-a)^{p-n}>0.$$
Quite similarly, 
$$i^p(-1)^n I_{n,p,a}=\Gamma (n-p)\,2^n(n+a-2j_{n,p,a})^{p-n}>0$$
if $a>n$ and $p$ is a non-integer in the interval $(-1,n)$. 
So, $I_{n,p,a}\ne0$ if $|a|>n$ and $p$ is a non-integer in the interval $(-1,n)$. 
By using the l'Hospital rule as mentioned in the question statement, one should likely get the same result when $|a|>n$ and $p$ is an integer in the interval $(-1,n)$. 

One can also consider the asymptotic behavior of $I_{n,p,a}$ as $n\to\infty$. Writing $e^{it}-e^{-it}$ (in the expression for $g_{n,a}(t)$) as $2i\sin t$ and then writing $\sin(\frac\pi2+k\pi+h)=(-1)^k\,e^{-h^2/(2+o(1))}$ for $k=0,1,\dots$ and $h\to0$, one sees that 
$$I_{n,p,a}\sim\frac{C_{p,a}}{\sqrt n}L(b/2,1/2,p+1)$$ 
as $n\to\infty$, where $C_{p,a}$ is manifestly nonzero and does not depend on $n$, $b=b_{n,a}$ equals $a$ or $a+1$ depending on whether $n$ is even or odd, and 
$$L(\lambda,c,s):=\sum_{k=0}^\infty\frac{\exp(2\pi i\lambda k)}{(k+c)^s}$$
is the Lerch zeta-function; see e.g. Wikipedia. 
It is now shown at positivity property of the Lerch zeta-function that $\Re L(b/2,c,p+1)>0$ for all real $b\ne0$, all real $c>0$, and all real $p>-1$. So, $I_{n,p,a}\ne0$ for large enough $n$. 
(If $b=0$, then $a\in\{-1,0\}$, in which case it is easy to see that $I_{n,p,a}\ne0$.) 
