The answer is no. Take $X \subset \mathbb{P}^4$ be a hyperplane section of $\mathbb{P}^1 \times \mathbb{P}^2 \subset \mathbb{P}^5$. So $X$ is a cubic ruled surface, I denote by $L$ the ruling.
Let $H$ be a generic hyperplane containing $L$. Then $H \cap X = L \cup D_1 \cup D_2$, where $D_1$ and $D_2$ are two disjoint lines of $X$ transverse to the ruling. Hence, if $x_1 = D_1 \cap L$ and $x_2 = D_2 \cap L$, then $H$ and $X$ have intersection multiplicity $2$ at $x_1$ and $x_2$.
Now, let $H_t$ be a family of such generic hyperplanes for $t \neq 0$ and such that $D_1$ colides to $D_2$ when $t \rightarrow 0$. Then, By Bezout, we have $H_0 \cap X = 2.D + L$.
Hence, the tangent cone of $X \cap H_0$ at $x_1 = x_2 =x$ is $2D$. It has degree $2$, so that the multiplicity of $H_0 \cap X$ at $x$ is $2$.
The subtle phenomenon here is that the tangency locus of $H_0$ along $X$ is non-reduced! It is equal to $D$ with an embedded point at $x$.
This paper : http://msp.org/pjm/2011/253-1/pjm-v253-n1-p01-p.pdf studies in details some phenomenon related to your question.