Your suspicion is correct. The following hypergraph $H$ provides a negative answer to your question. Let $V=\{0,1,\dots, 11\}$. Then $V=V_0\cup V_1\cup V_2$, where $V_0=\{0,1,2,3\}$, $V_1=\{4,5,6,7\}$, and $V_2=\{8,9,10,11\}$. Let $E(H)$ is a family of all three-element subsets $e$ of $V$, such that $|e\cap V_i|=1$ for each $i$ and the sum of elements of $e$ equals $0$ modulo $4$. By the construction, $H$ is a 3-partite 3-uniform hypergraph.

We claim that the matching width of $H$ equals $1$. Indeed, let $M$ be any non-empty matching in $H$. Suppose to the contary that $|M|=4$. Then $M$ covers each vertex of $H$ exactly once. Therefore the sum $S$ of vertices covered by $M$ equals $11\cdot 12/2=6$ modulo $4$. On the other hand, a sum of vertices covered by each edge of $M$ equals $0$ modulo $4$, and so does $S$, a contradiction. Therefore, $|M|\le 3$ and the following cases are possible.

1)) $|M|=1$. Then the unique edge of $M$ intersects itself, so $\rho(M)=1$.

2)) $|M|=2$. Let $M=\{\{a_0,a_1,a_2\}, \{b_0,b_1,b_2\}\}$, where $a_i, b_i\in V_i$ for each $i$. There exists a unique number $c\in V_2$ such that $a_0+b_1+c_2=0\pmod 4$. Then $\{a_0, b_1,c_2\}$ is an edge of $H$ intersecting each edge of $M$, so $\rho(M)=1$.

3)) $|M|=3$. Let $M=\{\{a_0,a_1,a_2\}, \{b_0,b_1,b_2\}, \{c_0,c_1,c_2\}\}$, where $a_i, b_i, c_i\in V_i$ for each $i$. There exist unique numbers $d_b, d_c\in V_2$ such that $a_0+b_1+d_b=0\pmod 4$ and $a_0+c_1+d_c=0\pmod 4$. Since $b_1\ne c_1\pmod 4$, $d_b\ne d_c$. Therefore the following cases are possible.

3.1)) $d_b\in \{a_2, b_2, c_2\}$. If $d_b=a_2$ then $b_1=a_1$, so $M$ is not a matching, a contradiction. If $d_b=b_2$ then $b_0=a_0$, so $M$ is not a matching, a contradiction. Thus $d_b=c_2$, and so $\{a_0, b_1, c_2\}$ is an edge of $H$ intersecting each edge of $M$, so $\rho(M)=1$.

3.2)) $d_c\in \{a_2, b_2, c_2\}$. If $d_c=a_2$ then $c_1=a_1$, so $M$ is not a matching, a contradiction. If $d_c=c_2$ then $c_0=a_0$, so $M$ is not a matching, a contradiction. Thus $d_b=b_2$, and so $\{a_0, c_1, b_2\}$ is an edge of $H$ intersecting each edge of $M$, so $\rho(M)=1$.

Thus $H$ has the matching width $1$.

On the other hand, we claim that $\tau(H)>3$. Indeed, let $Q$ be any three-element subset of $V$. The following cases are possible.

1)) There exists $V_i$ disjoint from $Q$. Let $V_j$ and $V_k$ be the remaining three-partite parts of $V$. Pick arbitrary numbers $v_i\in V_j\setminus Q$ and $v_k\in V_k\setminus Q $. There exists number $v_i\in V_i$ such that $v_i+v_j+v_k=0\pmod 4$. Then $\{v_i, v_j, v_k\}$ is an edge of $H$ disjoint from $Q$.

2)) $|Q\cap V_i|=1$ for each $i$. Pick any distinct numbers $v_0\in V_0\setminus Q$ and $v_1, u_1\in V_1\setminus Q$. There exist unique numbers $v_2, u_2\in V_2$ such that $v_0+v_1+v_2=0\pmod 4$ and $v_0+u_1+u_2=0\pmod 4$. Since $v_1\ne u_1\pmod 4$, $v_2\ne u_2$. Therefore the following cases are possible.

2.1)) $v_2\not\in Q$. Then $\{v_0, v_1, v_2\}$ is an edge of $H$ disjoint from $Q$.

2.2)) $u_2\not\in Q$. Then $\{v_0, u_1, u_2\}$ is an edge of $H$ disjoint from $Q$.