There is a $3\times 3 $ orthogonal matrix with all non zero entries.?

Gordon Pall found all rational orthogonal matrices, 3 by 3, in 1940, see PALL_PDF

Given an odd number $$ n = a^2 + b^2 + c^2 + d^2,$$ the matrix $$ \frac{1}{n} \begin{pmatrix} a^2 + b^2 - c^2 - d^2 & 2(-ad+bc) & 2(ac+bd) \\ 2(ad+bc) & a^2 - b^2 + c^2 - d^2 & 2(-ab+cd) \\ 2(-ac+bd) & 2(ab+cd) & a^2 - b^2 - c^2 + d^2 \end{pmatrix}$$ is orthogonal, and all rational orthogonal matrices can be written in this manner. This is formula (10) on page 755 of Pall's article, the second page of the pdf.

Right, so you want no zeroes in the matrix. Take, for example, $d=0,$ and nonzero $a,b,c$ which do not make a Pythagorean triple in any order. It would be enough to just add on the condition $c^2 \geq 1 + a^2 + b^2.$

Of course there is. Choose one vector with nonzero entries, say $(1,1,1)$, choose another one that is orthogonal to it, and that none of the three $2\times2$ minors they define vanishes, say $(1,2,-3)$. Now take their cross product to find a third vector orthogonal to the first two, and which has no zero entries by assumption; in this case it is $(-5,4,1)$. Now normalise everything (divide by square roots of $3,14,42$, respectively) and put the results as the columns of you matrix $A$.

Almost all unit vectors have a non-zero entry. Almost anything that you try should work. For example, if $A=(a, b, c)$ is a vector of length 1 with $a\neq -1$, you can verify that

$$ \begin{pmatrix} a & b & c \\ b & \frac {b^2-a-1}{a+1} & \frac {bc}{a+1} \\ c & \frac {bc}{a+1} & \frac {c^2-a-1}{a+1} \end{pmatrix}$$

describes an orthonormal set. This is the reflection of all the standard unit vectors across the line $e_1 +A$.