It's not NP-hard, unless P=NP. There is a polynomial-time algorithm to determine whether a single quadratic polynomial $f(x_1,\dots,x_n)$ has any zeros over $\mathbb{F}_2$, and if it does, to output one such zero. Write
$$f(x_1,\dots,x_n) = g(x_1,\dots,x_{n-1}) x_n + h(x_1,\dots,x_{n-1})$$
where $g$ is a degree-1 polynomial and $h$ is a degree-2 polynomial. Test whether $h(x_1,\dots,x_{n-1})$ has any zeroes over $\mathbb{F}_2$ (recursively).
If it does, then using that value of $x_1,\dots,x_{n-1}$ and setting $x_n=0$ gives a zero of $f(x_1,\dots,x_n)$.
Alternatively, if $h(x_1,\dots,x_{n-1})$ has no zeroes, then it follows that $h(x_1,\dots,x_{n-1})=1$ for all $x_1,\dots,x_{n-1}$, so we have
$$f(x_1,\dots,x_n) = g(x_1,\dots,x_{n-1}) x_n + 1.$$
You can check whether this has any zeros by setting $x_n=1$ and checking whether $g(x_1,\dots,x_{n-1}) + 1$ has any zeroes (which can be done using linear algebra, as $g$ has degree 1). (Obviously $x_n=0$ cannot lead to a zero of $f$ if $h$ is identically $1$, so there is no need to consider the case where $x_n=0$.)