While this post is **not** an answer the question, it does provide an observation which may help calculate $F(z;t)$
directly. 

It's not hard to see that

\begin{equation}
\begin{array}{lll}
\displaystyle H_t(1v) 
&\displaystyle = \, \big(n^2 - t \big) H_t(v)
&\text{if $|v|=n-1$} \\
\displaystyle H_t(2v) 
&\displaystyle = \, \big( 1 - t \big) \big( n^2 - t \big)  H_t(v)
&\text{if $|v|=n-2$} 
\end{array}
\end{equation}

and consequently

\begin{equation}
\langle H_t \rangle_n 
\ = \ 
{1 \over n} \big( n^2 - t \big)
\langle H_t \rangle_{n-1}
\ + \ {n-1 \over n} \big( 1 - t\big) \big(n^2 - t\big)
\langle H_t \rangle_{n-2}
\end{equation}
 
or, after setting 
$\omega(n) := {1 \over {n!}} \langle H_t \rangle_n$
that 

\begin{equation}
n^2 \omega(n) \ = \ 
\big(n^2 - t \big)\omega(n-1) \ + \
\big(1 - t \big)\big(n^2 - t \big) \omega(n-2)
\end{equation}

Of course $F(z;t) = \sum_{n \geq 0} \, \omega(n) z^n$ and it will satisfy the following second order homogeneous ODE in light of the linear recurrence:

\begin{equation}
\begin{array}{c}
\displaystyle A(z) {d^2 \over {dz^2}} F(z;t) \ + \
B(z) {d \over {dz}} F(z;t) \ + \
C(z) F(z;t) \ = \ 0 \\
\text{where} \\
\begin{array}{l}
A(z) \ = \ (1-t)z^3 + z^2 - z \\
B(z) \ = \ 5(1-t)z^2 + 3z - 1 \\
C(z) \ = \ (1-t)(4-t)z + (1-t)
\end{array}
\end{array}
\end{equation} 

I don't know how to solve this ODE but when 
$s=0$ one solution is $F(z;0) = (1 - z -z^2)$
which is the denominator of the partition function for the Fibonacci numbers. When $s=1$ one solution is
$F(z;1) = \int z^{-1} \, (z-1)^{-2} \, dz$ which can be computed by partial fractions. 

ines.