The 1-categorical question is well-defined, and its answer is no, in general. I will provide examples below.
Lemma 1. Let $\eta \colon X \to F$ be a map from $X$ to a fibrant object, initial among such maps. Then $\eta$ is a trivial cofibration. In particular, $F$ is a fibrant replacement of $X$.
Proof. $\require{AMScd}$ Factor $\eta$ as $\eta = pi$, where $i \colon X \to Z$ is a trivial cofibration and $p \colon Z \to F$ is a fibration. In particular, $Z$ is fibrant. By the universal property of $\eta$, there exists a (unique) map $\phi \colon F \to Z$ satisfying $i = \phi \eta$. The equality $p \phi \eta = p i = \eta$ then implies $p \phi = 1_F$, again by the universal property of $\eta$. The commutative diagram \begin{CD} X @= X @= X\\@V{\eta}VV @V{i}VV @V{\eta}VV\\ F @>{\phi}>> Z @>{p}>> F \end{CD} exhibits $\eta$ as a retract of $i$, hence a trivial cofibration. $\blacksquare$
Next, I will describe a model category in which the full subcategory of cofibrant objects is not coreflective. Taking the opposite model category, this provides a negative answer to the original question.
Let $R$ be a ring and consider $\mathrm{Ch}_{\geq 0}(R)$ the category of non-negatively graded chain complexes of (left) $R$-modules, equipped with the projective model structure. More precisely, the weak equivalences are quasi-isomorphisms, the fibrations are epimorphisms in positive degrees, and the cofibrations are degreewise monomorphisms with projective cokernel. In particular, a complex is cofibrant if and only if it is degreewise projective.
Observation 2. Let $M$ be an $R$-module, viewed as a complex concentrated in degree $0$. Then a chain map $f \colon C \to M$ is zero if and only if it induces the zero map on homology.
Proposition 3. Let $\epsilon \colon P \to M$ be a map from a cofibrant object to $M$, terminal among such maps. Then the following holds.
- Let $Q$ be a cofibrant chain complex and $g \colon Q \to P$ a chain map satisfying $\epsilon g = 0 \colon Q \to M$. Then $g$ is the zero map.
- The chain complex $P$ has trivial differential. In particular, its homology $H_n P = P_n$ is projective for every $n \geq 0$.
- $M$ is projective.
Proof. (1) follows from the universal property of $\epsilon$ and the equality $\epsilon g = 0 = \epsilon 0$. For (2), consider the chain complex $P_n[n-1]$ concentrated in degree $n-1$. The differential $d_n \colon P_n \to P_{n-1}$ defines a chain map $d_n \colon P_n[n-1] \to P$ which is null-homotopic. By Observation 2, we have $\epsilon d_n = 0$, and thus $d_n=0$. Now (3) follows from the isomorphism $H_0 P \cong M$, using Lemma 1. $\blacksquare$
Corollary 4. The full subcategory of cofibrant objects in $\mathrm{Ch}_{\geq 0}(R)$ is coreflective if and only if every $R$-module is projective (in which case every object of $\mathrm{Ch}_{\geq 0}(R)$ is cofibrant).