Let me expand on my comments at a bit more length.
First, there is a small inclarity in the question with the meaning of "restriction of the base-locus". Since the base locus is in general just a closed set, of possibly arbitrary codimension, it is not clear to me whether "restriction" is just meant to mean set-theoretic intersection, or something more sophisticated.
In any case, even in the simplest situation where the base locus is a codimension 1, the answer to both questions is no.
For the first version, let $X$ be $\mathbf P^2$ blown up in 9 points which are the intersection of two smooth cubics. Let $X \rightarrow \mathbf P^1$ be the elliptic fibration, and let $D$ be one of the exceptional divisors of $X \rightarrow \mathbf P^2$. Then $mD$ is fixed for all $m \geq 0$, but for a smooth fibre $F$ the restriction $D_{|F}$ is a point on an elliptic curve, so $2D_{|F}$ is basepoint free.
For the second version, fix an ample divisor $A$ on $X$ as above. For $D$ as before and for any $n \geq 1$, the divisor $A+nD$ is big. For $n$ sufficiently large, we have $(A + nD) \cdot D = A \cdot D -n <0$, so for such $n$ and sufficiently large $m$, the base locus of $m(A+nD)$ is again $D$. On the other hand, for any $n \geq 1$ the restriction of $A+nD$ to $F$ is basepoint free.