This is not an answer, but these are my thoughts so far and hopefully they will lead someone to a correct answer (hence the community wiki). My vague recollection is that the algebraic loop space only sees stuff "near" the constant loops, which is consistent with moonface's comments. I apologize if I am missunderstanding anything (I'm one of the struggling topologists). Basically I want to look at an example, which I think will elucidate the matter.
Let's take $X = \mathbb{G}_m$, the multiplicative group. Then $ \mathbb{G}_m(\mathbb{C}) = Spec \mathbb{C}[b, b^{-1}]$. As an analytic space I think this is just $\mathbb{C}^\times$, so on the topological side we get an interesting loop space. We have a fibration sequence,
$$\Omega \mathbb{C}^\times \to L\mathbb{C}^\times \to \mathbb{C}^\times $$
and since topologically $\mathbb{C}^\times \simeq S^1$, this shows that $\pi_0(L \mathbb{C}^\times) \cong \mathbb{Z}$. This is something that we should be able to detect if the algebraic version of the loop space is similar to the topological one, just count the number of components.
So what is the algebraic loop space in this case? Well, I guess by definition it is $Spec \; \mathbb{C}((t))[b,b^{-1}]$. Now remind me, how do we turn this into a space? and how many components does it have?
If you try to take the $\mathbb{C}$-points of it, i.e. homomorphisms, $$ \mathbb{C}((t))[b, b^{-1}] \to \mathbb{C}$$ don't you just get $\mathbb{C}^\times$? This seems to suggest that it is an infinitesimal thickening of the constant loops.
