Yes, this is true. More generally:$\newcommand{\CB}{\mathop{\sf{CB}}}\newcommand{\cbnorm}[1]{{\lVert#1\rVert}_{\sf cb}}$

**Proposition 1.** Let $X$, $Y$, $Z$ be operator spaces. Then composition of operators, viewed as a bilinear map $\CB(Y,Z) \times \CB(X,Y) \to \CB(X,Z)$, is jointly completely contractive.

I can't remember if this is proved explicitly in the book of Effros and Ruan, so I will attempt to give a leisurely explanation, which might not be the quickest or most direct approach, but is the one that I find easiest to reconstruct when needed.

I find it convenient to use a "curried" perspective. Namely: given vector spaces $E,F,G$ and a bilinear map $\beta:E\times F \to G$, define $T_\beta: E \to \mathop{\rm Lin}(F,G)$ by $T_\beta(x):y\mapsto \beta(x,y)$.

**Fact 2.** If $E$, $F$ and $G$ are operator spaces, $\beta$ is jointly completely contractive if and only if $T_\beta$ is a complete contraction from $E$ to $\CB(F,G)$.

(I will try to supply a reference for this equivalence when I get a chance to look in my copy of the Effros–Ruan book.)

Taking Fact 2 for granted, we see that to prove Proposition 1, it is enough to show that $\CB(Y,Z)\to \CB(\CB(X,Y),\CB(X,Z))$, $f \mapsto (g\mapsto f\circ g)$, is completely contractive. We do this in stages, by a kind of "boot-strap" process.

**Remark 3.** Let $E$ and $F$ be vector spaces, and note that for each $n\in {\mathbb N}$ there is a natural identification of $M_n \mathop{\rm Lin}(E,F)$ with $\mathop{\rm Lin}(E,M_nF)$. Then if $E$ and $F$ are operator spaces, this identification is a bijection between $M_n\CB(E,F)$ and $\CB(E,M_nF)$, and the **definition** of the operator space structure on $\CB(E,F)$ is that we define the norm on $M_n\CB(E,F)$ by transporting the norm on $\CB(E,M_nF)$ across this bijection.

**Lemma 4.** Let $E,F, G$ be operator spaces and let $h: F\to G$ be a complete contraction.

(i) For each $k\in\CB(E,F)$ we have $h\circ k \in \CB(E,G)$

(ii) The map $\CB(E,F) \to \CB(E,G)$ defined by $k\mapsto h\circ k$ is contractive.

*Proof.* Follows from the definition of the cb-norm (exercise).

**Corollary 5.** Let $X,Y,W$ be operator spaces and let $f:Y\to W$ be a complete contraction. Then the map $S: \CB(X,Y) \to \CB(X,W)$ defined by $g\mapsto f\circ g$ is *completely* contractive.

*Proof.* Let $n\in {\mathbb N}$; we need to show that the amplification $S^{(n)} : M_n\CB(X,Y) \to M_n\CB(Y,W)$ is contractive. This follows from Remark 3, by applying Lemma 4 with $h = f^{(n)} : M_nY \to M_n W$.

**Proof of Proposition 1.** Let $X, Y, Z$ be operator spaces and let $T:\CB(Y,Z) \to \CB(\CB(X,Y),\CB(X,Z))$ be given by composition of operators. As remarked above, we need to show that $T$ is completely contractive. So let $m\in {\mathbb N}$ and consider the amplification
$$
T^{(m)}: M_m \CB(Y,Z) \to M_m\CB(\CB(X,Y),\CB(Y,Z)).
$$
Using Remark 3 repeatedly, we may identify $T^{(m)}$ with the linear map
$$
\Theta:\CB(Y,M_m Z) \to \CB(\CB(X,Y),\CB(Y,M_mZ)
$$
defined by composition of operators. But now if $\cbnorm{f:Y\to M_mZ}\leq 1$, applying Corollary 5 with $W=M_mZ$ shows that $\cbnorm{\Theta(f): \CB(X,Y) \to \CB(Y,M_mZ)}\leq 1$. Thus $\Theta$ is a contraction, and we conclude that $\lVert T^{(m)}\rVert \leq 1$ as required.