No. Take
$$v_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \qquad v_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \qquad v_3 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}.$$
Then
$$P_1=v_1 v_1^T = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} \qquad P_2=v_2 v_2^T = \begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix} \qquad P_3=v_3 v_3^T = \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}$$
so
$$P_1 P_2 P_3 = \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}$$
which is not diagonalizable. (Replace $v_2$ by $\tfrac{1}{\sqrt{2}} \left[ \begin{smallmatrix} 1 \\ 1 \end{smallmatrix} \right]$ if you wanted orthonormal columns instead of just orthogonal ones.)

Regarding the random eigenvalues: For $(x,y,z)$ in $\mathbb{R}^3$, let $$f(x,y,z) = (x^2+y^2+z^2) \mathrm{Id}_3 - \begin{bmatrix} x\\y\\z \end{bmatrix} \begin{bmatrix} x&y&z \end{bmatrix}.$$
So $f(x,y,z)/(x^2+y^2+z^2)$ is orthogonal projection onto $\left[ \begin{smallmatrix} x\\y\\z \end{smallmatrix}\right]^{\perp}$.

I computed $f(1,0,0) f(u,v,w) f(x,y,z) f(1,0,0)$ for randomly chosen $(u,v,w,x,y,z)$ and, on my third try, I found the following example:
$$f(1,0,0) f(2,-3,-2) f(2,-3,1) f(1,0,0) = \begin{bmatrix}
0 & 0 & 0 \\
0 & 58 & -66 \\
0 & 33 & 143 \\
\end{bmatrix}.$$
The characteristic polynomial is
$$\lambda (\lambda^2 - 201 \lambda + 10472),$$
which has complex roots.