Yes. The equality can in fact be reached. Our strategy will be to produce a compact set $K_{d,r}\subseteq M_{d}(\mathbb{C})^{r}$ such that if $(X_1,\dots,X_r)\in K_{d,r}$, then $\rho(\Phi(X_1,\dots,X_r))=1$, and where
$$\rho_{2,d}(A_1,\dots,A_r)=\max\{\rho(A_1\otimes X_1+\dots+A_r\otimes X_r)\mid(X_1,\dots,X_r)\in K_{d,r}\}.$$
Let us go over a few definitions and facts to give some context to our construction. These facts can easily be found in John Watrous's 2018 book called The Theory of Quantum Information.
A mapping $\mathcal{E}:L(V)\rightarrow L(W)$ is said to be positive if whenever $X$ is positive semidefinite, $\mathcal{E}(X)$ is also positive semidefinite. A mapping $\mathcal{E}:L(V)\rightarrow L(W)$ is said to be completely positive if the mapping $\mathcal{E}\otimes 1_{U}:L(V\otimes U)\rightarrow L(W\otimes U)$ is positive for each finite dimensional complex Hilbert space $U$.
A mapping $\mathcal{E}:L(V)\rightarrow L(W)$ is said to be trace preserving if $\text{Tr}(\mathcal{E}(X))=\text{Tr}(X)$ for each $X\in L(V)$.
Proposition: Let $\mathcal{E}:L(V)\rightarrow L(W)$ be a linear operator.
$\mathcal{E}$ is completely positive if and only if there are $A_1,\dots,A_r$ where $\mathcal{E}=\Phi(X_1,\dots,X_r)$.
If $\mathcal{E}$ is defined by letting $\mathcal{E}(X)=A_1XB_1^*+\dots+A_rXB_r^*$, then $\mathcal{E}$ is trace preserving if and only if $A_1^*B_1+\dots+A_r^*B_r=1_V$.
A quantum channel is a completely positive trace preserving operator $\mathcal{E}:L(V)\rightarrow L(W)$. In quantum information theory, the quantum channels are the main morphisms between quantum states.
Proposition: If $\mathcal{E}$ is a quantum channel, then $\rho(\mathcal{E})=1$.
Let $K_{d,r}$ be the collection of all tuples $(X_1,\dots,X_r)\in M_{d}(\mathbb{C})^r$ such that $X_1^*X_1+\dots+X_r^*X_r=1_d$. Said differently, $K_{d,r}$ is the collection of all tuples $(X_1,\dots,X_r)\in M_{d}(\mathbb{C})^r$ such that
$\Phi(X_1,\dots,X_r)$ is a quantum channel. The set $K_{d,r}$ is clearly a closed set, and $K_{d,r}$ is bounded since $d=\text{Tr}(X_1^*X_1+\dots+X_r^*X_r)=\|X_1\|_2^2+\dots+\|X_r\|_2^2$ where $\|\cdot\|_2$ denotes the Frobenius norm, so $K_{d,r}$ is compact.
Lemma: Let $A_1,\dots,A_r\in L(V)$. Suppose that there is no $x\in V\setminus\{0\}$ with $A_1x=\dots=A_rx=0$. Furthermore, suppose that there is no subspace $W\subseteq V$ with $W\neq\{0\},W\neq V$, and $W=A_1[W]+\dots+A_r[W]$. Then there is a $\lambda>0$ along with a positive definite $P$ with $\Phi(A_1,\dots,A_r)(P)=\lambda P$.
Proof: Now, let $\mathcal{Q}$ be the collection of all positive semidefinite matrices in $L(V)$ with trace $1$, and let $F:\mathcal{Q}\rightarrow\mathcal{Q}$ be the mapping defined by letting $$F(P)=\frac{\Phi(A_1,\dots,A_r)(P)}{\text{Tr}\big(\Phi(A_1,\dots,A_r)(P)\big)}.$$ Then $\mathcal{Q}$ is convex, and $F$ is a continuous bijection, so by the Brouwer fixed point theorem, there is some $P\in\mathcal{Q}$ with
$F(P)=P$. Therefore, we have $\Phi(A_1,\dots,A_r)(P)=\lambda P$ for some positive $\lambda$.
Now,
$$\text{Im}(P)=\text{Im}(\lambda P)=\text{Im}(A_1PA_1^*+\dots+A_rPA_r^*)$$
$$=\text{Im}(A_1PA_1^*)+\dots+\text{Im}(A_rPA_r^*)=\text{Im}(A_1P)+\dots+\text{Im}(A_rP)$$
$$=A_1[\text{Im}(P)]+\dots+A_r[\text{Im}(P)].$$
This is only possible if $\text{Im}(P)=V$. $\square$
Let $O_{d,r}$ be the collection of all $(X_1,\dots,X_r)\in M_{d}(\mathbb{C})^{r}$ where $\Phi(X_1,\dots,X_r)$ is not nilpotent. Let $E_{d,r}$ be the collection of all $(X_1,\dots,X_r)\in M_d(\mathbb{C})^r$ where there is no subspace $W\subseteq\mathbb{C}^d$ with
$W=X_1^*[W]+\dots+X_r^*[W]$ and $W\neq\{0\},W\neq\mathbb{C}^d$ and where there is no $x$ with $X_j^*x=0$ for $1\leq j\leq r$.
The sets $O_{d,r},E_{d,r}$ are dense subsets of $M_{d}(\mathbb{C})^{r}$ with $E_{d,r}\subseteq O_{d,r}$.
Proposition: Suppose that $(X_1,\dots,X_r)\in E_{d,r}$. Then there is an invertible matrix $B$ and some positive number $\lambda$ with $(\lambda BX_1B^{-1},\dots,\lambda BX_rB^{-1})\in K_{d,r}$.
Proof: Suppose that $B$ is invertible and $\lambda$ is positive. Then $(\lambda BX_1B^{-1},\dots,\lambda BX_rB^{-1})\in K_{d,r}$ if and only if
$$\sum_{k=1}^{r}\lambda^2 (B^{-1})^*X_k^*B^*BX_k^*B^{-1}=I$$ if and only if
$$\sum_{k=1}^{r}\lambda^2 X_k^*B^*BX_k=B^*B$$.
Therefore, there are $\lambda,B$ with $(\lambda BX_1B^{-1},\dots,\lambda BX_rB^{-1})\in K_{d,r}$ if and only if there is a positive $\mu$ and a positive definite $P$ with $$\mu\Phi(X_1^*,\dots,X_k^*)P=P.$$
On the other hand, the existence of such a positive definite $P$ and positive $\mu$ is guaranteed by the above lemma. $\square$
Now, define a mapping $G_{A_1,\dots,A_r}:O_{d,r}\rightarrow\mathbb{R}$ by letting
$$G_{A_1,\dots,A_r}(X_1,\dots,X_r)=\frac{\rho(A_1\otimes X_1+\dots+A_r\otimes X_r)}{\rho(\Phi(X_1,\dots,X_r))^{1/2}}.$$ Since the mapping $G$ is continuous, we have
$$\rho_{2,d}(A_1,\dots,A_r)=\sup\{G_{A_1,\dots,A_r}(X_1,\dots,X_r)\mid (X_1,\dots,X_r)\in E_{d,r}\}.$$
Since $$G_{A_1,\dots,A_r}(X_1,\dots,X_r)=G_{A_1,\dots,A_r}(\lambda BX_1B^{-1},\dots,\lambda BX_rB^{-1})$$ whenever $B$ is invertible and $\lambda$ is a non-zero complex number, by the above proposition, we know that
$$\{G_{A_1,\dots,A_r}(X_1,\dots,X_r)\mid (X_1,\dots,X_r)\in E_{d,r}\}$$
$$=\{G_{A_1,\dots,A_r}(X_1,\dots,X_r)\mid (X_1,\dots,X_r)\in K_{d,r}\}.$$
Therefore, since $K_{d,r}$ is compact, there is some $(Z_1,\dots,Z_r)\in K_{d,r}$ with
$$G_{A_1,\dots,A_r}(Z_1,\dots,Z_r)=\max\{G_{A_1,\dots,A_r}(X_1,\dots,X_r)\mid (X_1,\dots,X_r)\in K_{d,r}\}$$
$$=\rho_{2,d}(A_1,\dots,A_r).$$
I have ran computations that maximize $G_{A_1,\dots,A_r}$, and in these computations the maximum seems to actually be reached.
