Yes, the canonical model structure is unique. The uniqueness of the canonical model structure on $Cat$ was nicely exposited by Chris Schommer-Pries on the Secret Blogging Seminar back in the day. Let's go through and mimic the proof there.
Explicit description of the canonical model structure:
Let $(C,W,F)$ denote the cofibrations, weak equivalences, and fibrations of the canonical model structure. The cofibrations are generated by $\emptyset \to pt$, along with $\Sigma X \to \Sigma Y$ for every map $X \to Y$ in $Top_\Delta$, where $\Sigma X$ is the $Top_\Delta$-enriched category with two objects $0,1$ and $Hom(0,0) = Hom(1,1) = pt$, $Hom(0,1) = X$, and $Hom(1,0) = \emptyset$. Therefore
$C$ consists of the injective-on-objects enriched functors.
$C \cap W$ consists of the injective-on-objects enriched equivalences. So it is generated by an inclusion $pt \to E$ where $E$ is the walking isomorphism.
$W$ consists of the enriched equivalences.
$W \cap F$ consists of the surjective-on-objects enriched equivalences.
$F$ consists of the isofibrations.
Uniqueness of this model structure:
Let $(C',W,F')$ be the cofibrations, weak equivalences, and fibrations of a model structure on $Cat_{Top_\Delta}$ with the same weak equivalences as the canonical model structure. We'll use "cofibration", "$\hookrightarrow$", "fibration", and $\twoheadrightarrow$ in the sense of this model structure, and "canonical cofibration" and "canonical fibration" to refer to the notions from the canonical model structure.
$\emptyset \hookrightarrow pt \in C'$.
For there must be some nonempty cofibrant object $X$, and then $0 \to pt$ is a retract of $\emptyset \hookrightarrow X$.
$C' \supseteq C$, $C' \cap W \supseteq C \cap W$, $F' \cap W \subseteq F \cap W$, $F' \subseteq F$.
For since $\emptyset \to pt \in C'$, we have that every $ f \in F' \cap W$ is surjective on objects. Since $f$ is also an enriched equivalence. we have $f \in F \cap W$, i.e. $F' \cap W \subseteq F \cap W$. The other inclusions follow formally.
If $C' \neq C$, then $E \to pt \in C' \cap W$ (where again $E = (0 \cong 1)$ is the walking isomorphism).
If $C' \neq C$, then by (2), $C' \not \subseteq C$, so there is a cofibration $A \hookrightarrow B \in C' \setminus C$ which is not injective on objects. Pick $x,y \in A$ which map to the same object in $B$. There is a unique enriched functor $A \to E$ sending $x$ to 0 and $y$ to 1. Then by pushout we obtain a cofibration $E \hookrightarrow E \cup_A B$ sending 0 and 1 to the same point. The codiscretification map $E \cup_A B \hookrightarrow (E\cup_A B)^\flat$ is injective on objects, i.e. a canonical cofibration, and hence a cofibration. Composing, we obtain a cofibration $E \hookrightarrow E \cup_A B \hookrightarrow (E\cup_A B)^\flat$ sending 0 and 1 to the same point, and sending the isomorphism between them to the identity. So $E \hookrightarrow pt$ is a retract of this map and hence also a cofibration. Since $E \to pt \in W$ is an enriched equivalence of categories, the claim follows.
If $C \neq C'$, then every fibrant object $X$ has no non-identity isomorphisms.
$X \to pt$ must lift against $E \to pt$ by (3).
$C = C'$, i.e. the model structures agree.
Every object must be equivalent to a fibrant object. But enriched equivalences preserve the property of having objects with non-identity automorphisms. So because there exist $Top_\Delta$-enriched categories with objects with non-identity automorphisms, this contradicts (4) if $C \neq C'$.
This argument generalizes to any cartesian enriching category $V$ such that there exist objects in $V$-categories with nontrivial automorphisms. I'd like to say this includes any cartesian enriching category which is not a poset, but I'm not quite sure.
