No. Let $a,b,c,d$ be complex numbers of modulus $1$, chosen so that $a\neq c$, $b\neq d$, and $ab^*\neq cd^*$. Form the column vectors $v=(1,a,b)^T$ and $w=(1,c,d)^T$. Then the matrix $$ vv^* +ww^*=\begin{pmatrix} 2 & a^*+c^* & b^*+d^* \\ a+c & 2 & ab^*+cd^* \\ b+d & a^*b+c^*d & 2 \end{pmatrix} $$ has rank two, but all the off-diagonal entries have modulus strictly less than $2$.

No. Let $a,b,c,d$ be complex numbers of modulus $1$, chosen so that $a\neq c$, $b\neq d$, and $ab^*\neq cd^*$. Form the column vectors $v=(1,a,b)^T$ and $w=(1,c,d)^T$. Then the matrix $$ vv^* +ww^*=\begin{pmatrix} 2 & a^*+c^* & b^*+d^* \\ a+c & 2 & ab^*+cd^* \\ b+d & a^*b+c^*d & 2 \end{pmatrix} $$ has rank two, but all the off-diagonal entries have modulus strictly less than $2$.

EDIT: A simple example would be $$ \begin{pmatrix} 1 & \frac35 & \frac45 \\ \frac35 & 1 & 0 \\ \frac45 & 0 & 1\end{pmatrix} $$ where one can replace $(\frac35, \frac45)$ with any pair $(a,b)$ such that $|a|^2+|b|^2=1$.