If you mean by an oblique cylinder a surface obtained by shearing a cylinder, then the answer is in general NO. Its cross-section perpendicular to the 'axial direction' will be an ellipse, which means that this surface is an elliptic cylinder.
Indeed, consider a cylinder $x^2 + y^2 = 1$ of radius $1$ and let us shear it to $x$-direction. Then with the shearing angle $\alpha$, the resulting surface is given by the equation $(x-z\tan\alpha)^2 + y^2 = 1$. Applying the rotation $(x, y, z) = (x'\cos\alpha + z'\sin\alpha, y', -x'\sin\alpha + z'\cos\alpha)$ so that the axial direction is now the $z'$-direction, the equation reduces to
$$ \frac{(x')^2}{\cos^2\alpha} + (y')^2 = 1, $$
which is indeed an elliptic cylinder. It is even obvious from the graph if we apply extreme degree of shearing: If $\tan\alpha = 3$, then
I will work this out assuming that the curve was supposed to lie in the plane $x+y+z = 3$, because in $\mathbb{R}^3$, the orthogonal complement of the associated plane through the origin has dimension $1$.
Notice that if the curve were supposed to lie in the $xy$-plane instead, we could set
$$
(x(t), y(t)) = 4 \cos(5t) (\cos t, \sin t) = 4 \cos(5t)[\cos t \, \textbf{e}_1 + \sin t \textbf{e}_2 ].
$$
This function gives a rose with $5$ petals because it is $\pi$-periodic, and there are precisely $5$ values of $t$ in $[0, \pi)$ for which $|\cos(5t)| = 1$. (Note that it is $\pi$-periodic, because $(\cos(t + \pi), \sin (t + \pi)) = -(\cos t, \sin t)$ and $\cos(5(t + \pi)) = \cos(5t + 5 \pi) = \cos(5t+ \pi) = - \cos(5t)$.)
This easier problem suggests a way forward in the case of the plane $x+y+z=3$: if we find an orthonormal basis $\{ \textbf{w}_1, \textbf{w}_2 \}$for the orthogonal complement of $(1,1,1)$ so that $\textbf{w}_1$ points in the direction of $(0,0,3) - (1,1,1)$, then we can replace $\textbf{e}_1$ and $\textbf{e}_2$ by the new basis vectors and translate by $(1,1,1)$ to give
$$
(x(t), y(t), z(t) ) = (1,1,1) + 4 \cos(5t) [\cos t \ \textbf{w}_1 + \sin \, \textbf{w}_2].
$$
Note that if we plug in $t = 0$, then we get
$$
(x(0), y(0), z(0)) = (1,1,1) + 4 \textbf{w}_1,
$$
so there is a petal in the direction of $\textbf{w}_1$, which is the direction of the line from the center $(1,1,1)$ to the point $(0,0,3)$.
Best Answer
The algebraic equation of a right circular cylinder with radius $r$ and axis $a$ (a unit vector) is given by
$ (p - p_0)^T (I - a a ^T) (p - p_0) = r^2 $
where $p_0$ is a point on the axis. To find the parametric equation, define a new basis for $\mathbb{R}^3$ having its $z'$-axis along the axis $a$. Let $u_1$ and $u_2$ be unit vectors that perpendicular to $a$ and to each other. Now express $p$ and $p_0$ in this basis as follows. First define the rotation matrix
$R = [u_1, u_2, a]$
Then define $q$ and $q_0$ by
$ p = R q $ and $p_0 = R q_0 $
Plug these into the equation of the cylinder
$ (q - q_0)^T R^T (I - a a ^T ) R (q - q_0) = r^2 $
Now $R^T (I - a a^T ) R = I - (R^T a) (R^T a)^T $
and $R^T a = \begin{bmatrix} 0 \\0 \\ 1 \end{bmatrix} $
Hence, $Q = R^T (I - a a ^T) R = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 0 \end{bmatrix} $
Thus the equation in terms of $q$ reduces to
$ (q_1 - q_{01} )^2 + (q_2 - q_{02})^2 = r^2 $
whose parametrization is
$q_1 = q_{01} + r \cos \theta $
$q_2 = q_{02} + r \sin \theta $
$q_3 = t $
where $t$ is arbitrary.
Finally, the parametrization of $p$ is
$ p = R q = R \begin{bmatrix} q_{01} + r \cos \theta\\ q_{02} + r \sin \theta\\ t \end{bmatrix} $