6. $\mathrm{C}$

In quadrilateral $A B C D$, point $D$ is not in the plane $A B C$, and $A C$ is connected.
By the property of a trihedral angle that the sum of two face angles is greater than the third face angle, we have
$$
\angle D A B<\angle B A C+\angle C A D, \angle B C D<\angle B C A+\angle D C A \text {. }
$$

Adding the two inequalities, we get
$$
\angle D A B+\angle B C D<\angle B A C+\angle C A D+\angle B C A+\angle D C A \text {. }
$$

Adding $\angle A B C+\angle C D A$ to both sides, we have
$$
\begin{array}{l}
\angle A B C+\angle B C D+\angle C D A+\angle D A B<\angle B A C+\angle C A D+\angle B C A+\angle D C A+\angle A B C+\angle C D A \\
=(\angle A B C+\angle B C A+\angle B A C)+(\angle D C A+\angle C D A+\angle C A D)=180^{\circ}+180^{\circ}=360^{\circ}
\end{array}
$$
6. In the spatial quadrilateral $ABCD$, points $A, B, C, D$ are not coplanar. The sum of the four interior angles is denoted as $\theta=\angle ABC + \angle BCD + \angle CDA + \angle DAB$, then
A. $\theta>360^{\circ}$
B. $\theta=360^{\circ}$
C. $\theta<360^{\circ}$
D. Cannot be determined