Denote $q=1-p$, write $[x^a]f(x)$ for a coefficient of $x^a$ in the polynomial (or Laurent polynomial) $f(x)$. We have $$F(n)=[1](px+q)^n(px^{-1}+q)^n=[1]((p^2+q^2)+qp(x+x^{-1}))^n=\\=(2\pi)^{-1}\int_{0}^{2\pi}((p^2+q^2)+qp(e^{it}+e^{-it}))^ndt.$$
The function $(p^2+q^2)+qp(e^{it}+e^{-it})=p^2+q^2+2pq\cos t$ is non-negative and its values do not exceed 1, thus the result.

I use the useful relation $[1]\varphi(x)=(2\pi)^{-1}\int_0^{2\pi} \varphi(e^{it})dt$ for any Laurent polynomial $\varphi$.

UPD. For the updated problem, use the same integral representation for $p=\sqrt{p_1p_2},q=\sqrt{q_1q_2} $. We need to check that $p+q\leqslant 1$, this follows from $p\leqslant (p_1+p_2)/2, $

$q\leqslant (q_1+q_2)/2. $

$$
\sum\limits_{i = 0}^k {C_k^ip_2^iq_2^{k - i}C_{n - k}^{k - i}p_3^{k - i}q_3^{\left( {n - k} \right) - \left( {k - i} \right)}}=[t^k](tp_2+q_2)^k(tp_3+q_3)^{n-k}
$$

About updated function $F(n)$. Denoting $h(z)=z^{-1}p_1(zp_2+q_2)+q_1(zp_3+q_3)$ we have $F(n)=[1](h(z))^n$, thus
$$
F(n)=\frac1{2\pi i}\int h(z)^ndz/z,$$
where integral is taken by any circuit around 0. It is natural to take the circuit with real value of $h(z)=p_1p_2+q_1q_3+q_1p_3 z+p_1q_2 z^{-1}$. The natural choice is the contour $z=\sqrt{\frac{p_1q_2}{q_1p_3}}e^{i\theta}$, we get $h(z)=p_1p_2+q_1q_3+2\sqrt{p_1q_1q_2p_3}\cos \theta$. We have $|h(z)|\leqslant p_1p_2+q_1q_3+p_1q_2+q_1p_3=p_1+q_1=1$, thus everything works nicely if we have $h(z)\geqslant 0$. This is not always the case. For example, if $p_2=q_3=0$ we get $h(z)=2\sqrt{p_1q_1}\cos \theta$ and the integral equals 0 for odd values of $n$ and is positive for even values of $n$. Needless to say, this could be seen from the very definition of $F(n)$, but for what it worth I leave here the general explanation.