This doesn't solve the question : it shows that a non-degenerate lattice always contains a primitive rank 2 sublattice with the required property regarding norms ... but the sublattice found is definite. Maybe a modification of the argument would yield the desired result, so I post it.

Without loss of generality we may assume $\Lambda$ is a $3$-dimensional maximal lattice. 

Let us write $B_N=\{v\in\Lambda, (v,v)\leq N\}$.

The positive definite case seems easy : for a given $N$, the ball $B_N$ in $\Lambda$ contains only a finite number of vectors, and there certainly exists a $2$-dimensional lattice that doesn't intersect $B_N-\{0\}$. 

The lorentzian case is a little subtler since then $B_N$ is not finite anymore. Here is an argument : let us see $\Lambda$ as a lattice in the standard Lorentzian space $\mathbf R^3$ with the form $q(x,y,z)=-x^2+y^2+z^2$ and let us choose the embedding so that the intersection of $\Lambda-\{0\}$ and the hyperplane $P$ defined by the equation $x=0$ is empty. The idea is that the set of vectors of $B_N$ that lie near $P$ is finite, while there are infinitely many hyperplanes of $\Lambda$ that are as near as you want from $P$. 

More precisely : let $x_N$ be the minimum of $\vert x\vert $ on $B_N-\{0\}$ (the chosen embedding implies $x_N>0$),  let $B_N'$ be the finite set of vectors of the form $(\pm x_N,y,z)$ in $B_N$ . Let $y_N$ (resp. $z_N$) be the maximum of $y$ (resp. $z$) on $B'_N$. 

Finally let $C$ be the set of vectors $(a,b,c)$ satisfying $\vert b\vert \leq \vert \frac{x_N}{3y_N} a\vert $ and $\vert c\vert\leq \vert \frac{x_N}{3z_N} a\vert$. This is a cone with a non-empty interior in the lorentzian space. Thus its intersection with $\Lambda-\{0\}$ is non-empty. Let us pick a vector $v=(a,b,c)$ therein, with $a\neq 0$. Then for a vector $(x,y,z)$ of the ortogonal complement of $v$, the inequalities $\vert y\vert\leq y_N$ and $\vert z\vert\leq z_N$ imply the inequality $x<x_n$. Thus we have $v^\perp\cap B_N=\{0\}$, and $v^\perp\cap\Lambda$ is a solution.