This post is **almost** an answer to the question. It provides a
method to directly calculate $F(z;t)$ and bypasses the obstacles 
related to computing the expectation values $\langle E_k \rangle_n$.

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 for general values
of the parameter $t$. 
Nevertheless, by construction, $F(z;0)$ 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}

and one can check directly that $(1 - z -z^2)^{-1}$
is indeed 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. However, any fibonacci word $u$ with
$|u| > 0$ must contain a box $\Box \in u$ with $\mathrm{h}(\Box)=1$
and so the statistic $H_1(u)$ must vanish.
Combinatorial realities thus force us to select the constant
solution $F(z;1) = 1$.
 


The recurrence above is clearly implemented by the
tridiagonal determinants. Specifically $\Bbb{f}_t(n)$ 
equals the following $n \times n$ tridiagonal determinant:


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

So the problem of finding $F(z;t)$ can be viewed as a special
case of the general problem of evaluating generating functions
of the sort

\begin{equation}
\sum_{n \geq 0} \, \det (T_n) \, z^n 
\end{equation}

where $T_n$ is the $n \times n$ leading, principal 
submatrix of a infinite $\Bbb{N} \times \Bbb{N}$
tridiagonal matrix $T$. As of yet, I don't know 
enough of the theory (presumably results in the study
of orthogonal functions and/or continued fractions) to
be able to answer. Any help would be appreciated. 

ines.