It is an interesting problem which is related to some recent work of mine. The reason for why I started to work on similar problems is because connections to a problem of Ramachandra on Dirichlet polynomials, connections to the nordic school of Hardy classes of Dirichlet series (Hedenmalm, Saksman, Seip, Olsen, Olofsson, Lindqvist and others), as well as universality questions for zeta-functions and their properties on the line Re(s)=1.
While my papers are not quite finished and I have 3 more papers related to similar problems that should be finished soon and put on arXiv I have put two early preprints on my homepage see On a problem of Ramachandra and approximation of functions by Dirichlet polynomials with bounded coefficients and On generalized Hardy classes of Dirichlet series. I have talked about some of these problems at analytic number theory conferences in India. Like in your paper I have considered Dirichlet series (it should be possible to obtain something like Theorem 2.1 in your paper by my method also, although I have not stated a direct analogue in my paper).
Now your problem in the question is rather easy for small $\omega$ so we will from now on assume that $\omega>1/2$. In fact if $\omega<1/2$, then $|f(0)|>1/2$ and $\int_0^K |\hat f(t)^2|dt \geq \min(1/10,K/10)$ (constants not chosen in an optimal way)
In my papers on Dirichlet series I have used a somewhat different method than you use in your paper, namely the Jensen inequality on the logarithmic integral in a half-plane. This method is applicable for the problem at hand. Lemma 7 in my paper ``On generalized Hardy classes of Dirichlet series'' can be used with $\sigma=0$ and $L(it)=\hat f(t)$ and we obtain $$\frac D \pi \int_{-\infty}^\infty \frac {\log^- |\hat f(t)|} {D^2+t^2} dt \leq \frac D \pi \int_{-\infty}^\infty \frac {\log^+ |\hat f (t)|} {D^2+t^2} dt - \log |\hat f(iD)|. $$ For similar results see also Koosis - The logarithmic integral. The reason why we can do this is that with the definition of the fourier-transform in your question it means that $ \hat f(z)$ will be a bounded analytic function in the half plane Im$(z) \geq 0$ and that $f(z) \to 0$ when Im$(z) \to \infty$.
Now in this case we also have that $\log^+ |\hat f (t)|=0$ since $ |\hat f (t)| \leq 1$. Thus the inequality simplifies to $$\frac D \pi \int_{-\infty}^\infty \frac {\log^- |\hat f(t)|} {D^2+t^2} dt \leq - \log |\hat f(iD)|.$$ It is not too difficult to see that for $\omega>1/2$ $$ |\hat f(i\omega)|= \left|\int_0^1 e^{i \phi(x)-\omega x} dx \right|>\frac {1} {10 \omega}. $$ (The constant $10$ not chosen optimally). Thus we can choose $D=\omega$ and it is clear that $$ \int_0^K \log^- |\hat f(t)| dt < \frac \pi {\omega} \left({\omega^2+K^2} \right) \frac {\omega} \pi \int_{-\infty}^\infty \frac {\log^- |\hat f(t)|} {\omega^2+t^2} dt $$ From these estimates we see that $$ \frac 1 K \int_0^K \log^- |\hat f(t)| dt< \frac {\pi(\omega^2+K^2)}{\omega K} \log (10 \omega). $$ Now we can use the Jensen inequality $$ \exp\left(\frac 1 K \int_0^K \log |\hat f(t)| \right)< \sqrt{\frac 1 K \int_0^K |\hat f(t)|^2 dt} $$ We get the lower bound $$ \left(\frac 1 {10 \omega} \right)^{2\pi (\omega^2+K^2)/(K \omega)} \leq \int_0^K |\hat f(t)|^2 dt $$ for $\omega>1/2$. If $\omega$ is of the same order as $K$ then the lower bound will be polynomial in $1/\omega$. While this does not give as good estimate as one would wish in this problem at least it give an explicit lower bound.