The answer is no.

Any surface of degree $d$ passing through the lines is of the form $l_1 Q_1+l_2Q_2$ where $Q_1,Q_2$ are homogeneous polynomials of degree $d-1$.

Since the line does not intersect the curve $C$, the curve $C$ is not in a plane passing through the line. Hence, neither $l_1$ neither $l_2$ vanishes on $C$. But this does not imply that $Q_1,Q_2$ vanish on $C$.

Edit: I gave an explicit counterample before, but there was a mistake in it, as observed by Mohan.

The answer is no in general.

Any surface of degree $d$ passing through the lines is of the form $l_1 Q_1+l_2Q_2$ where $Q_1,Q_2$ are homogeneous polynomials of degree $d-1$.

Since the line does not intersect the curve $C$, the curve $C$ is not in a plane passing through the line. Hence, neither $l_1$ neither $l_2$ vanishes on $C$. But this does not imply that $Q_1,Q_2$ vanish on $C$.

Let us give a precise example (sorry my previous one did not work because $C$ and $l$ were not empty, but this one is OK).

It suffices to choose a smooth cubic $S$ in $\mathbb{P}^3$, a line $l$ on it, and a curve $C$ on $S$ which does not intersect $l$ and which is not contained in any quadric. Then, the equation of $S$ if of the form $l_1Q_1+l_2Q_2$ where $Q_1,Q_2$ have degree $2$. BUT it is not possible to choose $Q_1,Q_2$ that vanish on $C$, since $C$ is not contained in any quadric.

It remains to see that such a curve $C$ exists; and in fact a general curve of $S$ will make the job. There exists a birational morphism $S\to \mathbb{P}^2$ which contracts $6$ lines of $S$ onto $6$ points of $\mathbb{P}^2$ in general position, and we can choose that $l$ is one of the lines contracted. Take then a curve of $\mathbb{P}^2$ of high degree (say $d>2$) not passing through any of the $6$ points. The preimage on $S$ gives a curve $C$ on $S$ which does not intersect $l$. Moreover, the degree of $C$ in $\mathbb{P}^3$ is the intersection with a hyperplane section, corresponding to a cubic of $\mathbb{P}^2$ though the six points, and is thus $3d>6$. Since $C$ is contained in a smooth cubic, and has degree $>6$, it is not contained in a quadric.

PS: The same argument does not work in smaller degree, i.e. if $S$ is a quadric. Indeed, the only curves in $S$ which do not intersect $l$ are lines, and are then contained in planes.

The answer is no.

Any surface of degree $d$ passing through the lines is of the form $l_1 Q_1+l_2Q_2$ where $Q_1,Q_2$ are homogeneous polynomials of degree $d-1$.

Since the line does not intersect the curve $C$, the curve $C$ is not in a plane passing through the line. Hence, neither $l_1$ neither $l_2$ vanishes on $C$. But this does not imply that $Q_1,Q_2$ vanish on $C$.

To have a precise counterexample, I take coordinates $[w:x:y:z]$ on $\mathbb{P}^3$, define $l_1=w$, $l_2=x$, and take $C$ to be the twisted cubic being the image of the morphism $\mathbb{P}^1\to \mathbb{P}^3$ given by $[u:v]\mapsto [uv^2:u^2v:u^3:v^3]$. Then $wy-x^2$ vanishes on $C\cup l$ but is not of the form $wR+xS$ with $R,S$ vanishing on $C$. Indeed, $R,S$ would be of degree $1$ and not polynomial of degree $1$ vanished on $C$.

The answer is no.

Any surface of degree $d$ passing through the lines is of the form $l_1 Q_1+l_2Q_2$ where $Q_1,Q_2$ are homogeneous polynomials of degree $d-1$.

Since the line does not intersect the curve $C$, the curve $C$ is not in a plane passing through the line. Hence, neither $l_1$ neither $l_2$ vanishes on $C$. But this does not imply that $Q_1,Q_2$ vanish on $C$.

Edit: I gave an explicit counterample before, but there was a mistake in it, as observed by Mohan.