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 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.
ines.