I like this question. What you describe is taking:
$${\bf \hat L}|l\rangle = \hat L_x|l\rangle {\bf \hat x} + \hat L_y \hat L_x|l\rangle {\bf \hat y} ... = {\bf \vec L}|l\rangle$$
which would lead to problems. I think an eigenvalue equation for a vector operator should be:
$${\bf \hat L}|l\rangle = \hat L_x|l\rangle {\bf \hat x} + \hat L_y|l\rangle {\bf \hat y} + \hat L_z|l\rangle {\bf \hat z} = L_x|l\rangle {\bf \hat x} + L_y|l\rangle {\bf \hat y} + L_z|l\rangle {\bf \hat z} = {\bf \vec L}|l>$$
where
$${\bf \hat L} = -i\hbar (y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y}){\bf \hat x} + (permutations) $$
is the operator and
$${\bf \vec L} = L_x {\bf \hat x}+ L_y {\bf \hat y} + L_z {\bf \hat z }$$
is an ordinary vector.
Now with:
$$ |l\rangle \equiv F_l(x, y, z)$$
the equation:
$$ {\bf \hat L}F_l(x, y, z) = {\bf \vec L}F_l(x, y, z)$$
doesn't have any non-trivial solutions: there are no eigenstates of completely known (non zero) angular momentum.
Regarding your question on "meaning": note that:
$$ L_z = -i\hbar\frac{\partial}{\partial\phi} $$
and eigenstates of that function look like:
$$ F(r, \theta, \phi) = R(r)\Theta(\theta)e^{\pm i m \phi} $$
which means that the function is invariant under rotations of $2\pi/m$.
For a wave-function to be an eigenstate of ${\bf \hat L}$ it would have to invariant under some non-trivial rotation about not just the $z$-axis, but the $x$ and $y$ axes also--which would mean it would have to be invariant under some rotation about any axis.
The only thing I can think of that can do that is a sphere, and this is, of course, the zero angular momentum eigenstate.