The answer is yes. For instance, consider $x_0 = -1$, $x_1 = -n$ and $x_2 = -n^2$, $n$ sufficiently large. All $x_i$'s are polynomials in $n$ and, for all large $n$ its sign is equal to the sign of the coefficient of the term of greatest degree. Let $a(x_n)$ such term. Then
$$\begin{aligned} a(x_0) &= -1, & a(x_1) &= -n, & a(x_2) &= -n^2, & a(x_3) &= -n^3, \\ a(x_4) &= n^3, & a(x_5) &= n^4, & a(x_6) &= -n^5, & a(x_7) &= n^7 \end{aligned}$$
Let's prove by induction that, for $k \ge 5$, $a(x_k) = a(x_{k-2})a(x_{k-3})$, which is equivalent to prove that the degree of $a(x_{k-1})$ is less than the degree of $a(x_{k-2}x_{k-3})$. This is true for $k = 5, 6, 7$. Suppose that it is true for all $k$ less than $m$. So the degree of $a(x_{m-1}) = a(x_{m-3})a(x_{m-4})$ is less than the degree of $a(x_{m-2})a(x_{m-3})$ because by the induction hypothesis the degree of $a(x_l)$ is less than the degree of $a(x_{l-1})$ and we're done.
So the sign of $x_k$ is equal to the sign of $x_{k-2}x_{k-3}$ for $k \ge 5$. Thus the signs of the sequence $x_k$ are
$$-, -, -, -, +, +, -, +, -, -, -, +, +, -, +, +, -, +, -, -, -, +, +, -, +, -, \dots$$
which are composed of several cycles of the form $-, -, -, +, +, -, +$. Since 4 out of the 7 terms in each cycle are negative, the result follows.