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. The first proof uses less machinery, but the second proves 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. This can clearly be generalized to any dimension, so every such sequence has a rational generating function and hence a constant-coefficient linear recurrence.

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.