Yes, the statement can be explicitly verified from the matrix representation of the spin operators acting on different spins. Acting on the spin-1/2 object, the spin operators read
$$S^x=\left(
\begin{array}{cc}
0 & \frac{1}{2} \\
\frac{1}{2} & 0 \\
\end{array}
\right), S^y=\left(
\begin{array}{cc}
0 & \frac{i}{2} \\
-\frac{i}{2} & 0 \\
\end{array}
\right), S^z=\left(
\begin{array}{cc}
-\frac{1}{2} & 0 \\
0 & \frac{1}{2} \\
\end{array}
\right).\qquad(1)$$
For spin-1 object, the spin operators read
$$S^x=\left(
\begin{array}{ccc}
0 & \frac{1}{\sqrt{2}} & 0 \\
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\
0 & \frac{1}{\sqrt{2}} & 0 \\
\end{array}
\right), S^y=\left(
\begin{array}{ccc}
0 & \frac{i}{\sqrt{2}} & 0 \\
-\frac{i}{\sqrt{2}} & 0 & \frac{i}{\sqrt{2}} \\
0 & -\frac{i}{\sqrt{2}} & 0 \\
\end{array}
\right), S^z=\left(
\begin{array}{ccc}
-1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{array}
\right).\qquad(2)$$
Using these operators, it is straight forward to verify that spin-1/2 object does not have quadrupole moment, but spin-1 object does have. For example, the quadrupole moment can be written in terms of the spin operators as $Q^{x^2-y^2}=(S^x)^2-(S^y)^2$ by definition. Plugging in Eq.(1) and complete the matrix multiplication, it can be verified that
$$Q^{x^2-y^2}=\left(
\begin{array}{cc}
0 & 0 \\
0 & 0 \\
\end{array}
\right),$$
meaning that the quadrupole moment vanishes for spin-1/2 object. However if we plug in Eq.(2), it can be found that
$$Q^{x^2-y^2}=\left(
\begin{array}{ccc}
0 & 0 & 1 \\
0 & 0 & 0 \\
1 & 0 & 0 \\
\end{array}
\right),$$
meaning that the quadrupole moment is non-vanishing for spin-1 object. Similar calculations can be done for other components of the quadrupole moment straightforwardly. It can be verified that all the five quadrupole moment operators are zero matrices in the spin-1/2 representation, thus explicitly proven that spin-1/2 object has no quadrupole moment.
The answer is basically an expansion of @Meng Cheng's comment.
