This depends on whether one insists on derived functors landing in the original model category (as is the case with modern approaches of Hinich and Dwyer–Hirschhorn–Kan–Smith), or in its homotopy category (as is the case with the classical approaches of Quillen, Grothendieck, etc.).
In the latter case it is indeed possible to define the left derived functor of a left Quillen functor in the original setting of (closed) model categories of Quillen, i.e., without the existence of functorial factorizations and small (co)limits, added later by Kan.
Given a left Quillen functor $$F\colon C→D,$$ pick a cofibrant replacement (not necessarily functorial) $Q_X$ and an acyclic fibration $q_X\colon Q_X→X$ for every object $X∈C$. Now construct the left derived functor $$\def\ldf{{\bf L}}\def\Ho{{\rm Ho}} \ldf F\colon C→\Ho(D)$$ as follows. Send an object $X∈C$ to $Q_X∈\Ho(D)$. Given a morphism $f\colon X→Y$ in $C$, factor the composition $f∘q_X\colon Q_X→Y$ through the acyclic fibration $q_Y\colon Q_Y→Y$ obtaining a map $g\colon Q_X→Q_Y$. Send $f$ to the morphism $$F(g)$$ in $\Ho(D)$.
To see that this construction yields a functor, consider composable morphisms $f\colon X→Y$, $g\colon Y→Z$ together with their chosen lifts $Q_f\colon Q_X→Q_Y$, $Q_g\colon Q_Y→Q_Z$, as well as the lift of the composition $Q_{gf}\colon Q_X→Q_Z$. Now $Q_g Q_f\colon Q_X→Q_Z$ is another lift for $gf\colon X→Z$. Any two lifts of the same morphism through an acyclic fibration are left homotopic (see, for example, Proposition 1.2.5(iv) in Hovey's book). Thus, the two lifts map to the same morphism in $\Ho(D)$.
An elementary argument then shows that this gives a functor, and a different choice of $q$ and $Q$ produces the same morphism in $\Ho(D)$, and therefore the same functor $C→\Ho(D)$.
Added: This is essentially the construction given by Quillen in Proposition I.4.1 of Homotopical Algebra, except that Quillen's construction starts by defining a functor $C→\Ho(C_c)$, where $C_c$ denotes the full subcategory of $C$ on cofibrant objects. This functor is then composed with $F$, using Ken Brown's lemma.