The answer to the first question is almost certainly NO. Start with a simple Lie algebra $\mathfrak{g}$ over $\mathbb{C}$ of type $E_6$. Using Bourbaki numbering for the vertices of its Dynkin diagram (with 2 assigned to the "extra" vertex), let $\mathfrak{g}'$ be the direct sum of the two rank 1 Lie subalgebras corresponding to the vertices 2 and 6. The subalgebra $\mathfrak{g}'$ has an involutive automorphism which interchanges these two copies of $\mathfrak{sl}_2$. But there seems to exist no extension to such an automorphism of $\mathfrak{g}$, essentially because the rank 1 subalgebras involved do not play interchangeable roles in the overall structure of $\mathfrak{g}$. (For example, their centralizers have different dimensions.)

Clearly this would require some work to make rigorous, but such a configuration of the pair $(\mathfrak{g}, \mathfrak{g}')$ indicates the kind of challenge you'd face here and also in your follow-up question.

[REVISED]

The answer to the first question is NO. It's difficult as a rule to show that an involutive automorphism of $\mathfrak{g'}$ fails to extend to $\mathfrak{g}$, so it may be better to consider first the possible automorphisms of a simple Lie algebra $\mathfrak{g}$ (up to conjugacy in the adjoint group). This is thoroughly worked out in the context of many other classification problems, in the classic 1978 book Differential Geometry, Lie Groups, and Symmetric Spaces by S. Helgason (reprinted as GSM 34, AMS, 2001). In Chapter X, $\S5$, Helgason studies automorphisms in the style of Victor Kac, which like other topics in this long chapter requires quite a bit of notation. But at the end of the section, Helgason explains in detail the possible automorphisms of $\mathfrak{g}$ having order 2, up to conjugacy. Here the fixed point subalgebra $\mathfrak{g}_0$ for $\mathfrak{g}$ of type $E_8$ can only be of type $D_8$ or else $A_1 \oplus E_7$.

This already shows that the pseudo-Levi subalgebra of type $D_8$ can't have a nontrivial involutive automorphism which extends to one in $E_8$, though $D_8$ does have a nontrivial graph automorphism. (The situation for an example such as $A_2$ embedded in $G_2$ is similar.)

As Paul Levy notes in his comments, there are other configurations (not involving simple subalgebras of simple Lie algebras) which also cause trouble here. I've included a couple of doubtful examples of this type in my earlier answer and comments, though as Paul comments it's not possible to rule out existence of some extensions by looking at centralizer dimensions alone.

All of this cautions against expecting to find many positive answers for pairs $(\mathfrak{g}, \mathfrak{g}')$.

[EARLIER TRY]

Start with a simple Lie algebra $\mathfrak{g}$ over $\mathbb{C}$ of type $E_6$. Using Bourbaki numbering for the vertices of its Dynkin diagram (with 2 assigned to the "extra" vertex), let $\mathfrak{g}'$ be the direct sum of the two rank 1 Lie subalgebras corresponding to the vertices 2 and 6. The subalgebra $\mathfrak{g}'$ has an involutive automorphism which interchanges these two copies of $\mathfrak{sl}_2$. But there seems to exist no extension to such an automorphism of $\mathfrak{g}$, essentially because the rank 1 subalgebras involved do not play interchangeable roles in the overall structure of $\mathfrak{g}$. (For example, their centralizers have different dimensions.) [EDIT: As Paul notes, this parenthetic comment is false.]

Clearly this would require some work to make rigorous, but such a configuration of the pair $(\mathfrak{g}, \mathfrak{g}')$ indicates the kind of challenge you'd face here and also in your follow-up question.

The answer to the first question is almost certainly NO. Start with a simple Lie algebra $\mathfrak{g}$ over $\mathbb{C}$ of type $E_6$. Using Bourbaki numbering for the vertices of its Dynkin diagram (with 2 assigned to the "extra" vertex), let $\mathfrak{g}'$ be the direct sum of the two rank 1 Lie subalgebras corresponding to the vertices 2 and 6. The subalgebra $\mathfrak{g}'$ has an involutive automorphism which interchanges these two copies of $\mathfrak{sl}_2$. But there seems to exist no extension to such an automorphism of $\mathfrak{g}$, essentially because the rank 1 subalgebras involved do not play interchangeable roles in the overall structure of $\mathfrak{g}$.

Clearly this would require some work to make rigorous, but such a configuration of the pair $(\mathfrak{g}, \mathfrak{g}')$ indicates the kind of challenge you'd face here and also in your follow-up question.

The answer to the first question is almost certainly NO. Start with a simple Lie algebra $\mathfrak{g}$ over $\mathbb{C}$ of type $E_6$. Using Bourbaki numbering for the vertices of its Dynkin diagram (with 2 assigned to the "extra" vertex), let $\mathfrak{g}'$ be the direct sum of the two rank 1 Lie subalgebras corresponding to the vertices 2 and 6. The subalgebra $\mathfrak{g}'$ has an involutive automorphism which interchanges these two copies of $\mathfrak{sl}_2$. But there seems to exist no extension to such an automorphism of $\mathfrak{g}$, essentially because the rank 1 subalgebras involved do not play interchangeable roles in the overall structure of $\mathfrak{g}$. (For example, their centralizers have different dimensions.)

Clearly this would require some work to make rigorous, but such a configuration of the pair $(\mathfrak{g}, \mathfrak{g}')$ indicates the kind of challenge you'd face here and also in your follow-up question.