Your sequence has generating function
$$
\frac{1+x+2x^2}{1-4x+x^2-2x^3}
$$
and satisfies the constant-coefficient linear recurrence
$$
b_{n+3} = 4b_{n+2}-b_{n+1}+2b_n.
$$

I'll give two quick proofs, but first I'll prove the 2D version with the same techniques. Both prove that the analogous problem in any number of dimensions always has a rational generating function and a constant-coefficient linear recurrence.

Consider the 2D case. Let $A(x)$ be the generating function for all valid walks, let $B(x)$ be the generating function for all valid walks that either end in $U$ or are length $0$, and let $C(x)$ be the generating function for all valid walks that end in $L$ or $R$. Then, these three formal power series satisfy the system of equations
$$
\left\{
\begin{array}{l}
A(x) = B(x) + C(x)\\
B(x) = 1 + xA(x)\\
C(x) = 2xB(x) + xC(x)
\end{array}
\right..
$$
The reasoning is as follows: every walk is either empty or ends in *something*, hence the first equation; every walk that ends in $U$ can be formed by taking *any* walk and appending a $U$, hence the second equation; every walk that ends in $L$ or $R$ can be formed by taking a walk that ends in $U$ or is empty and appending either $L$ or $R$ or by taking a walk that ends in $L$ or $R$ and appending the same symbol as the current last symbol. Solving this system gives
$$
A(x) = \frac{1 + x}{1-2x-x^2}
$$
which gives the constant-coefficient recurrence you mention.

Another solution can be formed with a two-state finite state machine. State 1 represents paths in which the next symbol can be anything, State 2 represents paths in which the next symbol is restricted because the previous symbol was $L$ or $R$. The start state is State 1 and both states are accepting. The transition matrix is
$$
\left[\begin{array}{cc}x&2x\\x&x\end{array}\right]
$$
and the transfer matrix method gives the same rational generating function.

For the 3D problem, you only need to one additional equation. Let $P(z)$ be the generating function for all valid walks, let $Q(z)$ be the generating function for all walks that end in $U$ or have length $0$, let $R(z)$ be the generating function for all walks that have a single direction locked in (i.e., since the last $U$, exactly one of $L,R,F,B$ has been seen), and let $S(z)$ be the generating function for all walks that have both directions locked in (i.e., since the last $U$, exactly one of $L$ or $R$ and exactly one of $F$ or $B$ have been seen). The system becomes
$$
\left\{\begin{array}{l}
P(x) = Q(x) + R(x) + S(x)\\
Q(x) = 1 + xP(x)\\
R(x) = 4xQ(x) + xR(x)\\
S(x) = 2xR(x) + 2xS(x)
\end{array}\right..
$$
The solution is the rational generating function I mention at the top of this post.

The second solution generalizes in a similar way. Now three states are needed: one where you are free to move in any direction, one in which one type of movement is locked ($L/R$ or $F/B$), and one in which both types of movement are locked.

These can clearly be generalized to any dimension, so every such sequence has a rational generating function and hence a constant-coefficient linear recurrence.