Are there examples of conditionally complete vector lattices that are not subsets of measurable functions (with order induced by cone of non-negative functions)?
I ask, because there are results in this direction: every Hilbert lattice is isomorph to some $L^2$ with order induced by pointwise inequalities, every lattice on a Hilbert space with self-dual positive cone is isomorph to $L^2$.
Are there similar results that say: a conditionally complete lattice with some extra structure is isomorph to some $L^2$?
There are conditionally order complete vector lattices (even conditionally order complete Banach lattices) which are not isomorphic to an $L^2$-space. For instance, for every measure space $(\Omega,\mu)$ and for $p \in [1,\infty)$ the space $L^p(\Omega,\mu)$ is conditionally order complete. If $(\Omega,\mu)$ is $\sigma$-finite, then $L^\infty(\Omega,\mu)$ is conditionally order complete, too.
If an $L^p$-space is infinite-dimensional and $p \not=2$, then it is not Banach space isomorphic to an $L^2$-space. Hence, it is not vector lattice isomorphic to an $L^2$-space, either (since each vector lattice isomorphism between two Banach lattices is automatically a Banach space isomorphism).
On the other hand, there are various representation theorems on how Archimedean vector lattices can be represented as sublattices of certain function spaces. An overview over many such representation thereoms can be found in Chapter 7 of the classical book "Riesz Spaces I" (1971) by Luxemburg and Zaanen.
In particular, every Archimedean vector lattice is a sublattice of a space of functions (but these functions might take the values $\infty$ and $-\infty$ on a "small" subset of their domain). This is the content of [op. cit., Section 44 in Chapter 7].