The answer is "Yes" though the construction (albeit very standard) is slightly non-obvious. 

First, we'll notice that for $f\in L^2$, you can make $f$ and $\hat f$ anything you want on $I=[-0.1,0.1]$, say. That is just because the product $PQ$ of the projections $Pf=f\chi_I$ and $Qf=(\hat f\chi_I)^\vee$ is a contraction in $L^2$, so to solve $Pf=Pg, Qf=Qh$, you just put $f=(g+h)-(Ph+Qg)+(QPh+PQg)-(PQPh+QPQg)+\dots$. Notice that if $g,h$ are real even, the series will consist of real even functions as well, so $f$ will be real even too. 

Now just choose some $C^\infty$ even real mollifier $\psi$ with support $(-0.01,0.01)$, $\int_{\mathbb R}\psi=1$, and positive $\hat\psi$. Solve the problem $f|I=0, \hat f|I=\frac{1}{\hat\psi}$ in $L^2$.

Take $F=f*\psi$. It will remain $0$ on $J=\frac 12I$ but we shall get $\hat F=\hat f\hat\psi|I=1$. and $\hat F$ will decay fast. Now convolve with $\psi$ on the Fourier side. You'll get a function $G\in S$ such that $G$ is real even, $G=0$ on $J$ and $\hat G=1$ on $J$. In particular,
$$
\int_{\mathbb R}G(t)=1,\quad \int_{\mathbb R}G(t)t^{2k}=0\text{ for }k\ge 1
$$ 

Since $G$ is even, we can say the same for the integrals from $0$ to $+\infty$, with $2G$ instead of $G$. It remains to change the variable, i.e., to take
$$
2G(t)=2t H(t^2)
$$ 
on $(0,+\infty)$ (since $G$ vanishes near the origin, that will result in $H\in S$ still). The function $H$ has exactly the properties you wanted.