It can be shown for any algebraic dependency $D(f(n-d), f(n-d+1), \dots, f(n+k))\equiv 0$, the polynomial $D(y_0,\dots,y_{d+k})$ belongs to the ideal generated by the irreducible factors of $k$ "consecutive" resultants $$\mathrm{Res}_n(Y_0, Y_1),\ \mathrm{Res}_n(Y_1, Y_2),\ \dots,\ \mathrm{Res}_n(Y_{k-1}, Y_k),$$ where $$Y_i := y_{i+d} - F(i+n,y_{i+d-1},\dots,y_i).$$ (Without loss of generality, we assume that $D(y_0,\dots,y_{d+k})$ depends on $y_{d+k}$.)
It further follows that we will have a counterexample to the conjecture if $\mathrm{Res}_n(Y_{k-1}, Y_k)$ (the only resultant depending on $y_{d+k}$) is irreducible and has degree in $y_{k+d}$ greater than 1.
Then, we have the following counterexample to both weak and strong conjectures with $d=2$: $$f(n) = (n+1)^2 f(n-1) + n^2 f(n-2),$$ where $\mathrm{Res}_n(Y_{k-1}, Y_k)$ (for whatever value of $k$) is irreducible and has degree of $y_{k+d}$ equals 2.