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 
$\Bbb{f}_t(n) := {1 \over {n!}} \langle H_t \rangle_n$


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

Of course $F(z;t) = \sum_{n \geq 0} \, \Bbb{f}_t(n) z^n$ 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 in general. 
By definition $F(z;0)$ must be 
the generating 
function for the Fibonacci numbers, i.e. 

\begin{equation} 
\begin{array}{l}
\displaystyle F(z;0) 
&\displaystyle = \ 1 + z + 2z^2 + 3z^3 + 5z^4 + \cdots \\
&\displaystyle = \ {1 \over {(1 - z -z^2)}}
\end{array}
\end{equation}

Indeed, one can check directly that $(1 - z -z^2)^{-1}$
is a solution to the ODE when $t=0$. When $t=1$ the ODE's 
general solution is $c_1 + c_2 \int z^{-1} \, (z-1)^{-2} \, dz$ where 
$c_1, c_2$ are constants; the later integral can be computed by partial fractions. 

My guess, is that $\Bbb{f}_t(n)$ is 
expressed as the following $n \times n$ tridiagonal determinant:


\begin{equation}
\det \begin{pmatrix}
\big( 1-t \big) &  {1 \over 4} \big(1-t \big) \big( 4 - t \big) & 0 & 0 & \\
-1 & {1 \over 4}\big(1 -t \big) & {1 \over 9} \big(1-t \big) \big( 9 - t \big) & 0 & \\
0 & -1 & {1 \over 9} \big(9 - t \big) & {1 \over 16} \big(1-t \big) \big( 16- t \big) & \\
0 & 0 & -1 & {1\over 16} \big(16 -t \big) & & \\
& & & & \ddots &
\end{pmatrix}
\end{equation}

ines.