The answer is **no**.

*Proof (Thomas Schick)*. The idea of the proof is due to Thomas Schick. I thank him for allowing me to reproduce it here. Let $\mathcal{A}:=C([0,1])\otimes\mathbb{M}_2$, where $\mathbb{M}_2$ is the $W^{\star}$-algebra of $2\times2$ matrices with entries in $\mathbb{C}$. Since the set of bounded Borel functions $B^{\infty}([0,1])$ is contained in $C([0,1])^{**}$, we canonically identify $B^{\infty}([0,1])\otimes\mathbb{M}_2$ with a $C^*$-subalgebra of $\mathcal{A}^{**}$. Let $F:=\{0\}$ and $F^c:=(0,1]$, then the characteristic functions $\chi_F$ and $\chi_{F^c}$ are Borel measurable. Define a partial isometry $u:=\chi_F\otimes E_{11}+\chi_{F^c}\otimes E_{12}\in\mathcal{A}^{**}$, where $E_{ij}\in\mathbb{M}_2$ is the canonical matrix unit, that is, it has $1$ in its $(i.j)$-entry and $0$ elsewhere. Then the orthogonal projection $p:=uu^{\star}=1\otimes E_{11}$ is open, whereas $q:=u^{\star}u=\chi_F\otimes E_{11}+\chi_{F^c}\otimes E_{22}$ is not. $\square$

Cf. Here is my related question.